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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