Projectile Motion Physics

Question

A soccer ball is kicked at 50 m/s at an angle of 37 degrees above the horizontal.

(a) Find the horizontal and vertical component of the initial velocity (2 pts)

(b) Determine the time it takes for the ball to return to the ground. (2 pts)

(c) Determine the range of the ball. (2 pts)

Answer 1

a/ Horizontal velocity = 50 cos 37 deg = 39.9 m/s

Vertical velocity = 50 sin 37 deg = 30.1 m/s

b/ Time of flight = 50 sin 37 / 4.9 = 6.14 sec

To see where this comes from you use the Position Equation

for the vertical direction

s(t) = s(0) + v( 0 vertical ) t + 1/2 a t^2 a = - g = - 9.8 m/s^2

0 = 0 + 50 sin 37 t - 4.9 t^2

c/ To find the Range or how far it goes use the Position Equation

for the horizontal direction

s(t) = s(0) + v( 0 horizontal ) t + 1/2 a t^2 .... a = 0

s( 6.14) = 0 + 50 cos 37 (6.14) + 0

s( 6.14) = 245.2 meters