Homework Help Question & Answers

A projectile is fired with an initial speed of 10 m/s. Find theangle of projection...

A projectile is fired with an initial speed of 10 m/s. Find the angle of projection such that the maximum height of the projectile is equal to 0.9 times its horizontal range.

3 0
Add a comment
✔ Recommended Answer
Answer #3

Sign Up to Unlock the answer FREE

Already have an account? Log in

ANSWER :


Initial speed = 10 m/s

Let the angle of projection be with the horizontal.

So, vertical component of initial speed = 10 sin


So, 


Total K.E. initially = 1/2 m (10)^2  (independent of ?

K.E. in the vertical direction = 1/2 m (10 sin ?)^2 

K.E. in the horizontal direction = 1/2 m (10 cos ?)^2


At maximum height, K.E in vertical direction = 0  (since velocity in the vertical direction is 0) 

but it gains P.E. by gaining the height h. 


So,


Gain of P.E = Loss of K.E.

=> m g h max = 1/2  m (10 sin ?)^2 

=> h max = (10 sin ? )^2  / 2g  =  (50/9.8) sin^2 ? = 5.102 sin^2 ? 

Time taken for attaining h max =  h max / average speed  = 5.102 sin^2 ? / (10 sin ? + 0)/2 = 1.0204 sin ?

Distance travelled horizontally when height is  h max = 10 cos ? * 1.0204 sin ? = 10.204 sin ? cos ?

Projectile reaches maximum height and comes back down and hits the ground. 

So, 

Total horizontal distance travelled 

= 2 * horizontal distance travelled when at h max

= 2 * 10.204 sin ? cos ?

= 20.408 sin ? cos ?


Given that :


h max = 0.9 total horizontal distance travelled

=> 5.102 sin^2 ? = 0.9 * 20.408 sin ? cos ?

=> sin ? / cos ? = 0.9 * 20.408 / 5.102 

=> tan ? = 3.6 

=> ? = 74.48 º  (with the horizontal)  (ANSWER).

answered by: Tulsiram Garg
Add a comment
Answer #1

Sign Up to Unlock the answer FREE

Already have an account? Log in

The Maximum height reached by the projectile is given by

Vy2 =Voy2+2ay*dY

Since final Velocity is zero

0=(10Sin(o))2+2*(-9.8)*dY

dY =(10Sin(o))2/19.6

The time of flight the projectile is twice time needed to reach the maximum height

t=2*(Vy-Voy)/a =2*(0-10sin(o))/(-9.8)

t=-20Sin(o)/9.8

the range of projectile is

dX =Voxt =(10Cos(o))*20Sin(o)/9.8

given

dY =0.9dX

(10Sin(o))2/19.6 =0.9*(10Cos(o))*20Sin(o)/9.8

tan(o) =3.6

o=tan-1(3.6) =74.5o

Add a comment
Answer #2

Sign Up to Unlock the answer FREE

Already have an account? Log in

ANSWER :


Initial speed = 10 m/s

Let the angle of projection be with the horizontal.

So, vertical component of initial speed = 10 sin


So, 


Total K.E. initially = 1/2 m (10)^2  (independent of ?

K.E. in the vertical direction = 1/2 m (10 sin ?)^2 

K.E. in the horizontal direction = 1/2 m (10 cos ?)^2


At maximum height, K.E in vertical direction = 0  (since velocity in the vertical direction is 0) 

but it gains P.E. by gaining the height h. 


So,


Gain of P.E = Loss of K.E.

=> m g h max = 1/2  m (10 sin ?)^2 

=> h max = (10 sin ? )^2  / 2g  =  (50/9.8) sin^2 ? = 5.102 sin^2 ? 

Time taken for attaining h max =  h max / average speed  = 5.102 sin^2 ? / (10 sin ? - 0)/2 = 1.0204 sin ?

Distance travelled horizontally when height is  h max = 10 cos ? * 1.0204 sin ? = 10.204 sin ? cos ?

Projectile reaches maximum height and comes back down and hits the ground. 

So, 

Total horizontal distance travelled 

= 2 * horizontal distance travelled when at h max

= 2 * 10.204 sin ? cos ?

= 20.408 sin ? cos ?


Given that :


h max = 0.9 total horizontal distance travelled

=> 5.102 sin^2 ? = 20.408 sin ? cos ?

=> sin ? / cos ? = 20.408 / 5.102 

=> tan ? = 4 

=> ? = 75.964 º  (with the horizontal)  (ANSWER).

answered by: Tulsiram Garg

> Please correct the last 4 lines as under :
=> 5.102 sin^2 𝛉 = 0.9 * 20.408 sin 𝛉 cos 𝛉
=> sin 𝛉 / cos 𝛉 = 0.9 * 20.408 / 5.102
=> tan 𝛉 = 3.6
=> 𝛉 = 74.48 º (with the horizontal) (ANSWER).

Tulsiram Garg Tue, Oct 26, 2021 8:29 AM

Add a comment
Know the answer?
Add Answer to:
A projectile is fired with an initial speed of 10 m/s. Find theangle of projection...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coin

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
Active Questions
ADVERTISEMENT