Question

What minimum speed is required for the ball to clear the 0.90-m-high net about 15.0 m from the server if the ball is "launched" from a height of h = 2.80 m ?

Answer 1

First, let us assign the variables

y = 0.90 m, x= 15 m,
`h_(0)` = 2.80 m

s= required

The vertical component of the trajectory is in uniformly accelerated motion. The equation is:


`y= v_(0,y) t+ (1)/(2) a t^(2) + h_(0)`, while the horizontal component is
`x= v_(0,x) t`. Also,
`v_(0,y) =0` since the object starts from rest (with respect to the downward motion). a is negative because it is moving downwards (a = -9.81 m/s^2). Substituting,


`0.9= 0+ (1)/(2) (-9.81m/ s^(2) ) ( (x)/( v_(0,x) ) )^(2) + h_(0)`


`0.9= 0+ (1)/(2) (-9.81m/ s^(2) ) ( (15)/( v_(0,x) ) )^(2) + 2.5`


`v_(0,x)= v_(0) =-24.1m/s`

The magnitude of v0, which is speed (s), is equal to 24.1 m/s