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

Step-by-step 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 -(gx^2)/(2u^2\cos^2\theta )`

For x=132 m

`y=132\tan 30-(9.8* 132^2)/(2* 44^2* (\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