Question

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net?

Answer 1

ball clears the net

Step-by-step explanation:

`v_(o)` = initial speed of launch of the ball = 20 ms^{-1}

`\theta` = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

`v_(ox)` = initial velocity =
`v_(o) Cos\theta = 20 Cos5 = 19.92 ms^(-1)`

`t` = time of travel

`X` = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

`X = v_(ox) t \\7 = v_(ox) t\\t = (7)/(v_(ox))` Eq-1

Consider the motion of the ball along the vertical direction

`v_(oy)` = initial velocity =
`v_(o) Sin\theta = 20 Sin5 = 1.74 ms^(-1)`

`t` = time of travel

`Y_(o)` = Initial position of the ball at the time of launch = 2 m

`Y` = Final position of the ball at time "t"

`a_(y)` = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

`Y = Y_(o) + v_(oy) t + (0.5) a_(y) t^(2)\\Y = 2 + (20 Sin5) ((7)/(20 Cos5)) + (0.5) (- 9.8) ((7)/(20 Cos5))^(2)\\Y = 2.00758 m\\`

Since Y > 1 m

hence the ball clears the net