Question

2. (25 pts) A tennis ball is hit with a speed of 21.5 m/s at angle of 15.0° above the horizontal from a height of 2.33 m. The top of the net is 1.07 m high. A) What is the farthest horizontal distance the tennis ball could be hit from the net and still make it over the net? B) What is spced of the ball as it clears the net if it is hit from the distance in part A?

Answer 1

here,

the initial speed , u = 21.5 m/s

theta = 15 degree

initial height , h0 = 2.33 m

height of net , h = 1.07 m

a)

let the horizontal distance be x

using equation of trajectory

(h - h0) = x * tan(theta) - g * x^2 /( 2 * u^2 * cos^2(theta))

( 1.07 - 2.33) = x * tan(15) - 9.81 * x^2 /( 2 * 21.5^2 * cos^2(15))

solving for x

x = 27.58 m

the horizontal distance is 27.58 m

b)

let the final vertical speed be vy

vy^2 - (u * sin(theta))^2 = - 2 * (h - h0) * g

vy^2 - ( 21.5 * sin(15))^2 = 2 * ( 2.33 - 1.07) * 9.81

solving for vy

vy = 7.46 m/s

vx = u * cos(theta) = 20.77 m/s

the final speed , v = sqrt(vx^2 + vy^2) = 22.1 m/s