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A batter hits a pitched ball when the center of the ball is 1.22 mabove the ground. The ball leaves the bat at an angle of 45with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 107 m.(a) Does the ball clear a 7.32 m-high fence that is 97.5 mhorizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

Short Answer

Expert verified
  1. The exact height of the ball at a given horizontal distance is 9.83 m from ground level.
  2. The distance between the ball鈥檚 center and fence is 2.51 m above the ground.

Step by step solution

01

Given information

It is given that, at the fence, the ball is at a height of 1.22 m above the ground,

the angle of projection is=45

The range of projectile isR=107mand

the horizontal distance isx=97.5m

02

Determining the concept

As ball is hit above the ground, take the point of fence as upside to ground.Horizontal range of the ball and the initial angle at which the ball is projected are given. So,kinematic equations can be used to find the initial velocity and time of the ball to reach a certain horizontal distance.

Required formulae are as follow:

The horizontal distance in kinematic equation can be written as,

x=v0t+12at2 (i)

The range of the projectile is given by,

R=v02sin2g (ii)

03

Determining the initial velocity

Let鈥檚 take the origin as the point of impact. To find the height of the ball at a given horizontal distance, find its initial velocity and time to reach that given horizontal distance.

Using equation (ii),

R=v02sin2gv0=Rgsin2v0=32.38m/s

04

Determining the time to reach the given horizontal distance

Using equation (i),

x=v0t+12at2

But, the time for horizontal distance and no acceleration in horizontal direction are needed.

So,

x=v0costt=xv0cost=4.258s

05

(a) Determining the height of the ball at a given horizontal distance

Now initial velocity and time to reach horizontal distance as well as the angle of projection are obtained. So by using 2nd kinematic equation, find the height of ball for that situation.

y=v0sint+12at2y=97.53-88.92y=8.61m

06

(b) Determining the distance between the ball’s center and fence

From the point of fence,

y=8.61+1.22=9.83m

So,theball is at 9.383 m height from the ground at 97.5 m horizontal distance.

And at 4.26 sec , the ball is 9.83-7.32=2.51mabove the fence.

Therefore, using the given conditions, the initial velocity oftheball can be found, and from this,thetime taken bytheball to reach some horizontal distance can be found. From all this information, the height of ball at that particular situation can be found.

Kinematic equations and equation of range of projectile, which is generally obtained from kinematic equations plays a main role in all the calculations. First of all, the value of initial velocity by using a range equation is obtained and that value will be helpful for finding the time of the ball to reach a certain horizontal distance.

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