A ball is thrown towards a wall 10 m away with a velocity of 40 m/s. At what angle must it be thrown if it is to enter a hole on the wall at a height of 7 m from the ground, neglecting wind effects (88.2°, 40.1°).

Question

Asked on June 22, 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 · 2026-06-22T09:05:51.211Z

Step 1: Identify the given values and the equations of projectile motion. Given: • Horizontal distance to the wall, x = 10 m • Height of the hole, y = 7 m • Initial velocity, v_0 = 40 m/s • Acceleration due to gravity, g = 9.8 m/s^2 (standard value) The equations for projectile motion are: Horizontal motion: x = v_0 () t Vertical motion: y = v_0 () t - (1)/(2)gt^2 Step 2: Express time t from the horizontal motion equation. From the horizontal motion equation, we can solve for t: t = (x)/(v_0 ()) Step 3: Substitute t into the vertical motion equation to get the trajectory equation. Substitute the expression for t into the vertical motion equation: y = v_0 () ((x)/(v_0 ())) - (1)/(2)g ((x)/(v_0 ()))^2 Simplify the equation: y = x (())/(()) - (gx^2)/(2v_0^2 ^2()) Using the trigonometric identities () = (())/(()) and (1)/(^2()) = ^2() = 1 + ^2(): y = x () - (gx^2)/(2v_0^2) (1 + ^2()) Step 4: Rearrange the trajectory equation into a quadratic equation for (). Let T = (). The equation becomes: y = xT - (gx^2)/(2v_0^2) (1 + T^2) Multiply by 2v_0^2 to clear the denominator: 2v_0^2 y = 2v_0^2 x T - gx^2 (1 + T^2) 2v_0^2 y = 2v_0^2 x T - gx^2 - gx^2 T^2 Rearrange into the standard quadratic form AT^2 + BT + C = 0: gx^2 T^2 - 2v_0^2 x T + (gx^2 + 2v_0^2 y) = 0 Step 5: Substitute the given values and solve the quadratic equation for T. Substitute x=10 m, y=7 m, v_0=40 m/s, and g=9.8 m/s^2: A = gx^2 = (9.8)(10)^2 = 9.8 × 100 = 980 B = -2v_0^2 x = -2(40)^2(10) = -2(1600)(10) = -32000 C = gx^2 + 2v_0^2 y = (9.8)(10)^2 + 2(40)^2(7) = 980 + 2(1600)(7) = 980 + 22400 = 23380 The quadratic equation is: 980 T^2 - 32000 T + 23380 = 0 Divide by 20 to simplify: 49 T^2 - 1600 T + 1169 = 0 Using the quadratic formula T = -B ± sqrt(B^2 - 4AC)2A: T = -(-1600) ± sqrt((-1600)^2 - 4(49)(1169))2(49) T = 1600 ± sqrt(2560000 - 229124)98 T = 1600 ± sqrt(2330876)98 T = (1600 ± 1526.7147)/(98) This gives two possible values for T: T_1 = (1600 + 1526.7147)/(98) = (3126.7147)/(98) ≈ 31.90525 T_2 = (1600 - 1526.7147)/(98) = (73.2853)/(98) ≈ 0.747809 Step 6: Calculate the angles _1 and _2. _1 = (T_1) = (31.90525) ≈ 88.20^ _2 = (T_2) = (0.747809) ≈ 36.78^ The two angles at which the ball can be thrown are approximately 88.20^ and 36.78^. The problem provides (88.2°, 40.1°) as expected values. Our first calculated angle matches exactly. The second calculated angle is close to 40.1^, with the difference likely due to rounding in the problem's provided value or a slightly different value of g used. The angles at which the ball must be thrown are: 88.20^ and 36.78^ That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-06-22T09:05:51.211Z

Step 1: Identify the given values and the equations of projectile motion. Given: • Horizontal distance to the wall, x = 10 m • Height of the hole, y = 7 m • Initial velocity, v_0 = 40 m/s • Acceleration due to gravity, g = 9.8 m/s^2 (standard value) The equations for projectile motion are: Horizontal motion: x = v_0 () t Vertical motion: y = v_0 () t - (1)/(2)gt^2 Step 2: Express time t from the horizontal motion equation. From the horizontal motion equation, we can solve for t: t = (x)/(v_0 ()) Step 3: Substitute t into the vertical motion equation to get the trajectory equation. Substitute the expression for t into the vertical motion equation: y = v_0 () ((x)/(v_0 ())) - (1)/(2)g ((x)/(v_0 ()))^2 Simplify the equation: y = x (())/(()) - (gx^2)/(2v_0^2 ^2()) Using the trigonometric identities () = (())/(()) and (1)/(^2()) = ^2() = 1 + ^2(): y = x () - (gx^2)/(2v_0^2) (1 + ^2()) Step 4: Rearrange the trajectory equation into a quadratic equation for (). Let T = (). The equation becomes: y = xT - (gx^2)/(2v_0^2) (1 + T^2) Multiply by 2v_0^2 to clear the denominator: 2v_0^2 y = 2v_0^2 x T - gx^2 (1 + T^2) 2v_0^2 y = 2v_0^2 x T - gx^2 - gx^2 T^2 Rearrange into the standard quadratic form AT^2 + BT + C = 0: gx^2 T^2 - 2v_0^2 x T + (gx^2 + 2v_0^2 y) = 0 Step 5: Substitute the given values and solve the quadratic equation for T. Substitute x=10 m, y=7 m, v_0=40 m/s, and g=9.8 m/s^2: A = gx^2 = (9.8)(10)^2 = 9.8 × 100 = 980 B = -2v_0^2 x = -2(40)^2(10) = -2(1600)(10) = -32000 C = gx^2 + 2v_0^2 y = (9.8)(10)^2 + 2(40)^2(7) = 980 + 2(1600)(7) = 980 + 22400 = 23380 The quadratic equation is: 980 T^2 - 32000 T + 23380 = 0 Divide by 20 to simplify: 49 T^2 - 1600 T + 1169 = 0 Using the quadratic formula T = -B ± sqrt(B^2 - 4AC)2A: T = -(-1600) ± sqrt((-1600)^2 - 4(49)(1169))2(49) T = 1600 ± sqrt(2560000 - 229124)98 T = 1600 ± sqrt(2330876)98 T = (1600 ± 1526.7147)/(98) This gives two possible values for T: T_1 = (1600 + 1526.7147)/(98) = (3126.7147)/(98) ≈ 31.90525 T_2 = (1600 - 1526.7147)/(98) = (73.2853)/(98) ≈ 0.747809 Step 6: Calculate the angles _1 and _2. _1 = (T_1) = (31.90525) ≈ 88.20^ _2 = (T_2) = (0.747809) ≈ 36.78^ The two angles at which the ball can be thrown are approximately 88.20^ and 36.78^. The problem provides (88.2°, 40.1°) as expected values. Our first calculated angle matches exactly. The second calculated angle is close to 40.1^, with the difference likely due to rounding in the problem's provided value or a slightly different value of g used. The angles at which the ball must be thrown are: 88.20^ and 36.78^ That's 2 down. 3 left today — send the next one.

A ball is thrown towards a wall 10 m away with a velocity of 40 m/s. At what angle must it be thrown if it is to enter a hole on the wall at a height of 7 m from the ground, neglecting wind effects (88.2°, 40.1°).

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A ball is thrown towards a wall 10 m away with a velocity of 40 m/s. At what angle must it be thrown if it is to enter a hole on the wall at a height of 7 m from the ground, neglecting wind effects (88.2°, 40.1°).

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