Question

A soccer ball is kicked with a speed of 35 m/s at an angle of 30° with respect to the ground and lands a certain distance away from the point it was kicked. If a second ball were kicked from the same place with a speed of 37 m/s, at what angle with respect to the ground could its initial velocity be such that it lands in the same place as the first ball?

Answer 1

θ = 25.4º

Step-by-step explanation:

For this exercise we can use the projectile launch equations, let's use the scope ratio

R = vo² sin 2θ / g

For the first ball

R1 = 35² 2 sin (2 30) /9.8

R1 = 108.25 m

For the second ball, they ask that the second ball fall in the same place, so the horizontal distance is the same

R2 = R1

Therefore we calculate the angle

sin 2θ = R1 g / vo²

sin 2θ = 108.25 9.8 / 37²

sin 2θ = 0.7749

2θ = sin⁻¹ (0.7749)

2θ = 50.8º

θ = 25.4º