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

$x = vcos\theta * t$

$y = v sin\theta * t - \frac{1}{2} gt^2$

solving above two equations we have

$y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}$

now here we will plug in all data

$0.9 = 122 tan51 - \frac{9.8 * 122^2}{2*v^2 * cos^251}$

$0.9 = 150.65 - \frac{184150.2}{v^2}$

$\frac{184150.2}{v^2} = 149.75$

$v = 35.1 m/s$

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