Posted by Sarah on Sunday, June 7, 2009 at 7:49pm.
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?

Physics  MathMate, Monday, June 8, 2009 at 1:06am
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_{v}^{2}  u_{v}^{2}
2*(9.81)*S = 20.13^{2}
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(0^{2}+u_{h}^{2})
=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.
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