Question

Question #88f6b

Answer 1

The ball will clear the wall only when the initial speed is more than about 19 m/s.

Explanation:

Let
`v_0` (m/s): initial speed of the ball
`v_(0x)` (m/s): initial speed along the horizontal direction
`v_(0y)` (m/s): initial speed along the vertical direction.

Then, it will take `t=32/v_(0x)=32/(v_0cos40°)=41.77/v_0` (s) to reach the wall.
At that time, the height of the ball is
`h=v_(0y)t-1/2g t^2`
`=v_0sin40° * 41.77/v_0 -1/2 * 9.807 * (41.77/v_0)^2`
`=26.85-8555/v_0^2` [m].

When the ball clears the wall, the inequation
`26.85-8555/v_0^2>3.5` must be satisfied.
`8555/v_0^2<23.35`
`8555<23.35v_0^2`
`v_0^2> 366.4`
`v_0>19.14` (m/s)