A ball is kicked with an initial velocity of 25 s at an angle of 35° to the ground. Calculate: (a) The maximum height reached. (b) The time of flight. (c) The range.

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Asked on September 7, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-09-07T01:41:51.929Z

here are the solutions for all the questions from the image. --- Question 5: A tennis ball is projected with a velocity of 20 m/s into the air at an angle of 30^ with respect to the horizontal axis. (i) Calculate the horizontal component of the velocity at the maximum height R. Step 1: Identify the horizontal component of initial velocity. The horizontal component of velocity (v_x) remains constant throughout the projectile's flight, assuming no air resistance. v_x = v_0 Step 2: Substitute the given values. v_x = 20 m/s × 30^ v_x = 20 m/s × sqrt(3)2 v_x ≈ 17.32 m/s The horizontal component of the velocity at maximum height is 17.32 m/s. (ii) Calculate the vertical component of the velocity at the maximum height R. Step 1: Understand velocity at maximum height. At the maximum height of a projectile's trajectory, the vertical component of its velocity momentarily becomes zero. The vertical component of the velocity at maximum height is 0 m/s. (iii) If the maximum horizontal distance reached is 50 m, calculate the time of flight. Step 1: Identify given values and the formula. Given: Maximum horizontal distance (Range), R = 50 m. Initial velocity, v_0 = 20 m/s. Launch angle, = 30^. The horizontal component of velocity is constant: v_x = v_0 . The time of flight (T) is related to the range by R = v_x T. Step 2: Calculate the horizontal component of velocity. v_x = v_0 = 20 m/s × 30^ v_x = 20 m/s × 0.866 v_x = 17.32 m/s Step 3: Calculate the time of flight. T = (R)/(v_x) T = 50 m17.32 m/s T ≈ 2.89 s The time of flight is 2.89 s. --- Question 6: A tennis ball rolls off a 1.5 m high table at an initial horizontal velocity of 2 m/s. (a) How long does it take for the ball to reach the floor? Step 1: Use the vertical motion equation. The vertical displacement is h = 1.5 m. The initial vertical velocity v_0y = 0 m/s. h = v_0yt + (1)/(2)gt^2 Step 2: Substitute the values. 1.5 m = (0 m/s)t + (1)/(2)(9.8 m/s^2)t^2 1.5 = 4.9t^2 Step 3: Solve for t. t^2 = (1.5)/(4.9) t = sqrt((1.5)/(4.9)) t ≈ 0.553 s The time taken to reach the floor is 0.55 s. (b) How far does the ball land from the table? Step 1: Use the horizontal motion equation. The horizontal distance (R) is given by R = v_0xt. Step 2: Substitute the values. Using the time t ≈ 0.553 s from part (a) and v_0x = 2 m/s. R = 2 m/s × 0.553 s R ≈ 1.11 m The ball lands approximately 1.11 m from the table. (c) What is the impact velocity of the ball? Step 1: Calculate the horizontal and vertical components of the impact velocity. The horizontal velocity (v_x) remains constant: v_x = v_0x = 2 m/s The vertical velocity (v_y) at impact is: v_y = v_0y + gt v_y = 0 m/s + (9.8 m/s^2)(0.553 s) v_y ≈ 5.42 m/s Step 2: Calculate the magnitude of the impact velocity. The impact velocity (v) is the resultant of its horizontal and vertical components: v = sqrt(v_x^2 + v_y^2) v = sqrt((2 m/s))^2 + (5.42 m/s)^2 v = sqrt(4 + 29.38) v = sqrt(33.38) v ≈ 5.78 m/s The impact velocity of the ball is approximately 5.78 m/s. --- Question 7: A ball is kicked with an initial velocity of 25 m/s at an angle of 35^ to the ground. (a) The maximum height reached. Step 1: Use the formula for maximum height (H). H = (v_0^2 ^2 )/(2g) Step 2: Substitute the values. Given v_0 = 25 m/s, = 35^, g = 9.8 m/s^2. 35^ ≈ 0.5736 H = (25 m/s)^2 (0.5736)^22 × 9.8 m/s^2 H = (625 × 0.3289)/(19.6) H = (205.5625)/(19.6) H ≈ 10.49 m The maximum height reached is 10.49 m. (b) The time of flight. Step 1: Use the formula for the time of flight (T). T = (2 v_0 )/(g) Step 2: Substitute the values. 35^ ≈ 0.5736 T = 2 × 25 m/s × 0.57369.8 m/s^2 T = (50 × 0.5736)/(9.8) T = (28.68)/(9.8) T ≈ 2.93 s The time of flight is 2.93 s. (c) The range. Step 1: Use the formula for the range (R). R = (v_0^2 (2))/(g) Step 2: Substitute the values. 2 = 2 × 35^ = 70^ 70^ ≈ 0.9397 R = (25 m/s)^2 (70^)9.8 m/s^2 R = (625 × 0.9397)/(9.8) R = (587.3125)/(9.8) R ≈ 59.93 m The range is 59.93 m. Drop the next question!

Accepted Answer · 2026-09-07T01:41:51.929Z

here are the solutions for all the questions from the image. --- Question 5: A tennis ball is projected with a velocity of 20 m/s into the air at an angle of 30^ with respect to the horizontal axis. (i) Calculate the horizontal component of the velocity at the maximum height R. Step 1: Identify the horizontal component of initial velocity. The horizontal component of velocity (v_x) remains constant throughout the projectile's flight, assuming no air resistance. v_x = v_0 Step 2: Substitute the given values. v_x = 20 m/s × 30^ v_x = 20 m/s × sqrt(3)2 v_x ≈ 17.32 m/s The horizontal component of the velocity at maximum height is 17.32 m/s. (ii) Calculate the vertical component of the velocity at the maximum height R. Step 1: Understand velocity at maximum height. At the maximum height of a projectile's trajectory, the vertical component of its velocity momentarily becomes zero. The vertical component of the velocity at maximum height is 0 m/s. (iii) If the maximum horizontal distance reached is 50 m, calculate the time of flight. Step 1: Identify given values and the formula. Given: Maximum horizontal distance (Range), R = 50 m. Initial velocity, v_0 = 20 m/s. Launch angle, = 30^. The horizontal component of velocity is constant: v_x = v_0 . The time of flight (T) is related to the range by R = v_x T. Step 2: Calculate the horizontal component of velocity. v_x = v_0 = 20 m/s × 30^ v_x = 20 m/s × 0.866 v_x = 17.32 m/s Step 3: Calculate the time of flight. T = (R)/(v_x) T = 50 m17.32 m/s T ≈ 2.89 s The time of flight is 2.89 s. --- Question 6: A tennis ball rolls off a 1.5 m high table at an initial horizontal velocity of 2 m/s. (a) How long does it take for the ball to reach the floor? Step 1: Use the vertical motion equation. The vertical displacement is h = 1.5 m. The initial vertical velocity v_0y = 0 m/s. h = v_0yt + (1)/(2)gt^2 Step 2: Substitute the values. 1.5 m = (0 m/s)t + (1)/(2)(9.8 m/s^2)t^2 1.5 = 4.9t^2 Step 3: Solve for t. t^2 = (1.5)/(4.9) t = sqrt((1.5)/(4.9)) t ≈ 0.553 s The time taken to reach the floor is 0.55 s. (b) How far does the ball land from the table? Step 1: Use the horizontal motion equation. The horizontal distance (R) is given by R = v_0xt. Step 2: Substitute the values. Using the time t ≈ 0.553 s from part (a) and v_0x = 2 m/s. R = 2 m/s × 0.553 s R ≈ 1.11 m The ball lands approximately 1.11 m from the table. (c) What is the impact velocity of the ball? Step 1: Calculate the horizontal and vertical components of the impact velocity. The horizontal velocity (v_x) remains constant: v_x = v_0x = 2 m/s The vertical velocity (v_y) at impact is: v_y = v_0y + gt v_y = 0 m/s + (9.8 m/s^2)(0.553 s) v_y ≈ 5.42 m/s Step 2: Calculate the magnitude of the impact velocity. The impact velocity (v) is the resultant of its horizontal and vertical components: v = sqrt(v_x^2 + v_y^2) v = sqrt((2 m/s))^2 + (5.42 m/s)^2 v = sqrt(4 + 29.38) v = sqrt(33.38) v ≈ 5.78 m/s The impact velocity of the ball is approximately 5.78 m/s. --- Question 7: A ball is kicked with an initial velocity of 25 m/s at an angle of 35^ to the ground. (a) The maximum height reached. Step 1: Use the formula for maximum height (H). H = (v_0^2 ^2 )/(2g) Step 2: Substitute the values. Given v_0 = 25 m/s, = 35^, g = 9.8 m/s^2. 35^ ≈ 0.5736 H = (25 m/s)^2 (0.5736)^22 × 9.8 m/s^2 H = (625 × 0.3289)/(19.6) H = (205.5625)/(19.6) H ≈ 10.49 m The maximum height reached is 10.49 m. (b) The time of flight. Step 1: Use the formula for the time of flight (T). T = (2 v_0 )/(g) Step 2: Substitute the values. 35^ ≈ 0.5736 T = 2 × 25 m/s × 0.57369.8 m/s^2 T = (50 × 0.5736)/(9.8) T = (28.68)/(9.8) T ≈ 2.93 s The time of flight is 2.93 s. (c) The range. Step 1: Use the formula for the range (R). R = (v_0^2 (2))/(g) Step 2: Substitute the values. 2 = 2 × 35^ = 70^ 70^ ≈ 0.9397 R = (25 m/s)^2 (70^)9.8 m/s^2 R = (625 × 0.9397)/(9.8) R = (587.3125)/(9.8) R ≈ 59.93 m The range is 59.93 m. Drop the next question!

A ball is kicked with an initial velocity of 25 s at an angle of 35° to the ground. Calculate: (a) The maximum height reached. (b) The time of flight. (c) The range.

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A ball is kicked with an initial velocity of 25 s at an angle of 35° to the ground. Calculate: (a) The maximum height reached. (b) The time of flight. (c) The range.

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