Question

A tennis player hits a ball at ground level, giving it an initial velocity of 26.0m/s at 55.0? above the horizontal.

What is the ball's acceleration at its highest point?

When this ball lands on the court, how far is it from the place where it was hit?

Answer #1

vx0 = 26*cos(55) = 14.91m/s; vy0 = 26.0*sin(55) = 21.29m/s

Max height occurs when vy = 0 so vy^2 = vy0^2 - 2*g*ymax ..so
ymax = vy0^2/2*g

= 21.29^2/(2*9.8) = 23.12m

Now use vy = vy0 - g*t ...so t = vy0/g = 21.29/9.8 = 2.17s

v at the top = vx only = vxo = 14.91m/s

a = -g = -9.8m/s^2

Now use y = y0 + vy0*t - 1/2*g*t^2 here y = y0 =0

so vy0*t -1/2*g*t^2 = 0 so t*(vy0 - 1/2*g*t) = 0 ..Therefore t=
2*vy0/g = 2*21.29/9.8 = **4.34s**

x = vx0*t = 14.91m/s*4.34 = **64.78m**

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