Serving at a speed of 155 km/h, a tennis player hits the ball at a height of 2.5

Question

Serving at a speed of 155 km/h, a tennis player hits the ball at a height of 2.57 m and an angle e below the horizontal. The player serves at a distance of 12.0 m from the net, which is 0.91 m high. What is the angle ? such that the ball just crosses the net? Submit Answer Tries 0/30 The out line of the service box is 6.40 m from the net. How far does the ball land from the out line of the service box? (If the ball lands outside the out line, enter your answer as a positive number. If inside, enter as a negative number.) Submit Answer Tries 0/30

Explanation / Answer

let v = 155 km/h = 155*5/18 = 43.06 m/s

x = 12 m

H = 2.57 m
h = 0.91 m

let t is the time taken for the ball to reach the net

use,

t = x/vx = 12/(43.06*cos(theta))

now in y-direction,

H - h = vy*t + (1/2)*g*t^2

2.57 - 0.91 = 43.06*sin(theta)*(12/(43.06*cos(theta)) + (1/2)*9.8*(12/(43.06*cos(theta))^2

1.66 = 12*tan(theta) + 4.9*(12/43.06)^2/(cos(theta))^2

==> theta= 6.06 degrees <<<<<<<----------------------Answer

b) let R is the total distance travelled in horizontal direction befpre hitting the ground.

let t is the time taken

2.57 = 43.06*sin(6.06)*t + (1/2)*9.8*t^2

==> t = 0.39617 s

so, R = vx*t

= 43.06*cos(6.06)*0.39617

= 16.96 m

so, the ball touches the ground at = 16.96 - (12 + 6.4)

-1.44 m <<<<<<<----------------------Answer

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