Question

Assuming the ball's initial velocity was 51 ∘ above the horizontal and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? assume that the ground was perfectly flat. express your answer using two significant figures.

Answer 1

horizontal distance of home run is 400 ft = 122 m

height of the home run is 3 ft = 0.9 m

now the angle of the hit is 51 degree

now we have equation of trajectory of the motion

`x = vcos\theta * t`

`y = v sin\theta * t - (1)/(2) gt^2`

solving above two equations we have

`y = xtan\theta - (gx^2)/(2v^2cos^2\theta)`

now here we will plug in all data

`0.9 = 122 tan51 - (9.8 * 122^2)/(2*v^2 * cos^251)`

`0.9 = 150.65 - (184150.2)/(v^2)`

`(184150.2)/(v^2) = 149.75`

`v = 35.1 m/s`

so the ball was hit with speed 35.1 m/s from the ground