Question

Recall that the equation for horizontal distance "h" in feet of a projectile with initial velocity v0 and initial angle theta is given by h=(v0^2/16)sin theta cos theta. a.) Assume the initial velocity is 60 (feet/second). What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet? b.) Assume the initial velocity is 60 feet/second. What is the maximum horizontal distance possible and at what angle does this occur?

Answer 1

A

100 = (60^2 / 16) sin theta cos theta
100 = 225 sin theta cos theta
sin theta cos theta = 100/224 = 0.4444
1/2 sin 2theta = 0.4444
sin 2theta = 0.8888
2 * theta = 62.73 degrees
theta = 31.37 degrees

Answer 2

A). 31.43°

B). 225 feet

Explanation:

A) From the given equation for horizontal distance h of a projectile is given by

`h=(v²_(0))/(16)sin\theta cos\theta`

If initial velocity v0 = 60 ft/sec and horizontal distance = 100 feet

Then the equation will be
`h=(v_(0)^(2))/(16)sin\theta cos\theta`

`100=(60^(2))/(32)sin2\theta`

`100=(3600)/(32)sin2\theta`

`sin2\theta =32* 100/3600=0.89`

`2\theta =62.87`

`\theta =31.43`

B). If v0 = 60 feet per second

Then we have to calculate the maximum horizontal distance and angle.

Since we know to cover the maximum distance in a projectile motion an object should be thrown at 45°.

Therefore the equation formed will be

`h=(60^(2))/(32)sin2* 45`

`=(3600)/(32)sin90`

`=112.5* 1=225`

h = 112.5 feet