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?

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

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

    B. Max height, S
    consider vertical direction only
    uv = 20.13 (initial)
    vv = 0 (final, at highest point)
    a= -g = -9.81 m/s/s
    2*a*S = vv2 - uv2
    2*(-9.81)*S = -20.132
    S=20.65 m

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

    D. At highest point,
    velocity =sqrt(02+uh2)
    =13.07 m/s (horizontal)
    = -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, Sh
    =uh * T
    = 13.07 m/s * 4.1 s
    = 53.6 m.

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  2. The solving is so cute and great with proper explanation

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  3. Thanks

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