Question

Can I know how to solve this physics question?

Answer 1

Assuming that the ball was `120\ cm` above the ground when hit.

The initial velocity has horizontal and vertical components as

Horizontal: `v_(0x) = 40.0\ cos 26.0^@\ ms^-1`
Vertical: `v_(0y) = 40.0\ sin 26.0^@\ ms^-1`

Both components are orthogonal to each other, therefore these can be treated separately.

Time `t` taken by the ball to travel `110\ m` in the horizontal direction. Ignoring air resistance, horizontal velocity of ball remains constant.

`:. t = 110/(40.0\ cos26.0^@)\ s`

To calculate the height of the ball at time `t` can be found using the kinematic expression

`y = y_0 + ut + 1/2g t^2`

Noting that gravity acts in a direction opposite to the initial direction of vertical velocity, inserting various values we get

`y = 1.20 + (40.0\ sin26.0^@)xx110/(40.0\ cos26.0^@) +`
`1/2xx(-9.81)xx(110/(40.0\ cos26.0^@))^2`
`=>y=1.20+53.650-45.981`
`=>y = 8.93\ m`

Fielder can reach `=3\ m` above the ground.
`:.` Height of ball passing above glove `= 8.93-3=5.93\ m`