Question

A baseball leaves the bat with a speed of 44.0 m/s and an angle of 30.0° above the horizontal. A 5.0-m-high fence is located at a horizontal distance of 132 m from the point where the ball is struck. Assuming the ball leaves the bat 1.0 m above ground level, by how much does the ball clear the fence?

Answer 1

Answer:13.41 m

Explanation:

Given

initial velocity (u)=44 m/s

launch angle \theta =30[/tex]

height of fence=5 m

Horizontal distance of fence=132 m

Ball leaves the launch with a height of 1 m

Trajectory of Projectile is given by

$y=x\tan \theta -\frac{gx^2}{2u^2\cos^2\theta }$

For x=132 m

$y=132\tan 30-\frac{9.8\times 132^2}{2\times 44^2\times (\cos 30)^2}$

$y=76.21-58.8=17.41 m$

ball is already 1 m so net height of ball w.r.t ground is

Y=17.41+1=18.41

so ball will clear the fence by a distance of 18.41-5=13.41 m