Question

A tennis player hits a ball at ground level, giving it an initial velocity of 24 m/s at 57 degrees above the horizontal. (a) what are the horizontal and verticla components of the ball;s initial velocity? (b) how high above the ground does the ball go? (c) How long does it take the ball to reach its maximum height? (d) What are the ball;s velocity and acceleration at its highest point? (e) for how long a time is the ball in the air? (f) When this ball lands on the court, how far is it from the place where it was hit?

Answer 1

u = initial velocity = 24 m/s
θ = angle = 57 degrees

A. components
horizontal component, u_h
= u cos(θ)
= 24*cos(57)
= 13.07 m/s
vertical component, u_v
= u sin(θ)
= 24*sin(57)
= 20.13 m/s

B. Max height, S
consider vertical direction only
u_v = 20.13 (initial)
v_v = 0 (final, at highest point)
a= -g = -9.81 m/s/s
2*a*S = v_v² - u_v²
2*(-9.81)*S = -20.13²
S=20.65 m

C. Time to reach max. height, t
Consider vertical direction only:
v=u+at
0 = 20.13 + (-9.81)*t
t = 20.13/9.81 = 2.05 s.

D. At highest point,
velocity =sqrt(0²+u_h²)
=u_h
=13.07 m/s (horizontal)
acceleration
= -g
= -9.81 m/s/s

E. Ball in the air, T
Time = 2t = 2*2.05 s = 4.1 s.

F. Horizontal distance travelled, S_h
=u_h * T
= 13.07 m/s * 4.1 s
= 53.6 m.