A projectile is launched from ground level at an angle of 35° above the horizontal. It passes through a point that is 15 meters high and 30 meters away horizontally from the launch point. Using only the kinematic equations of projectile motion, find the initial velocity of the projectile.

Step 1: Write down the kinematic equations for horizontal and vertical motion. The horizontal displacement x is given by: x = (v_0 ) t The vertical displacement y is given by: y = (v_0 ) t - (1)/(2) g t^2 We are given x = 30 m, y = 15 m, = 35^, and g = 9.8 m/s^2. We need to find v_0. Step 2: Express time t from the horizontal motion equation. From the horizontal motion equation: 30 = (v_0 35^) t t = (30)/(v_0 35^) Step 3: Substitute the expression for t into the vertical motion equation. 15 = (v_0 35^) ( (30)/(v_0 35^) ) - (1)/(2) (9.8) ( (30)/(v_0 35^) )^2 Step 4: Simplify the equation and solve for v_0. 15 = 30 ( 35^)/( 35^) - 4.9 (30^2)/((v_0 35^)^2) 15 = 30 35^ - (4.9 × 900)/(v_0^2 ^2 35^) 15 = 30 35^ - (4410)/(v_0^2 ^2 35^) Rearrange to solve for v_0^2: (4410)/(v_0^2 ^2 35^) = 30 35^ - 15 v_0^2 = (4410)/((30 35^ - 15) ^2 35^) Now, calculate the numerical values: 35^ ≈ 0.7002 35^ ≈ 0.81915 ^2 35^ ≈ (0.81915)^2 ≈ 0.6709 Substitute these values: v_0^2 = (4410)/((30 × 0.7002 - 15) × 0.6709) v_0^2 = (4410)/((21.006 - 15) × 0.6709) v_0^2 = (4410)/((6.006) × 0.6709) v_0^2 = (4410)/(4.0306) v_0^2 ≈ 1094.1 v_0 = sqrt(1094.1) v_0 ≈ 33.077 m/s The initial velocity of the projectile is approximately 33.1 m/s. Send me the next one 📸