A ball is to be shot from level ground with a certain speed. Figure 4-45 shows the range Rit will have versus the launch angle θ0. The value ofθ0determines the flight time; let tmaxrepresent the maximum flight time. What is the least speed the ball will have during its flight ifθ0is chosen such that flight time isrole="math" localid="1660897865281" 0.500tmax?

Short Answer

Expert verified

Least speed of the ball at given condition 42.0 m/s

Step by step solution

01

Given

From the graph we can say that maximum range of projectile 240 m is which can be reaching by it only when launching angle is45°.

02

Understanding the concept

First, find the time to reach the maximum height for the object. We know that at the maximum height the object has only horizontal velocity component. Its vertical will be zero. So,the angle for half of maximum height can be found. After that from the graph we can conclude the initial velocity of object.

Formulae:

Time of flight,

t=2v0sinθ0g …(i)

Range of the projectile,

R=v02sin2θg …(ii)

The x component of the velocity,

vox=vocosθ …(iii)

03

Calculate the angle of projection

We know that that time to reach the maximum height is

t=2v0sinθ0g

Time required to reach the maximum height by the object whenθ=90°is,

tmaximum=2v0g

Therefore, half of the maximum time of flight is given by,

tmaximum2=v0g

That means,

sinθ0=12θ0=30°

04

Calculate the least speed the ball

We know from graph is range is 240 m, and angle is 45°. So, we can find its initial velocity.

R=v02sin2θg

After plugging values, we getv0=48.5m/s

And at maximum height object has only horizontal velocity component

vox=v0cosθ

So, at half maximum height object have velocity will,

vox=(48.5m/s)cos(30°)=42.0m/s

Therefore, the least speed of the ball at given condition.

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