Home
Class 11
PHYSICS
The veritical height of the projectile a...

The veritical height of the projectile at time t is given by `y= 4t -t ^(2),` and the horizontal distance covered is given by `x = 3t.` What is the angle of projection with the horizontal ?

A

`tan^(-1)3//5`

B

`tan^(-3)4//5`

C

`tan^(-1)4//3`

D

`tan^(-1)3//4`

Text Solution

Verified by Experts

The correct Answer is:
C

`y=4t-t^(2),x=3t`
`V_(y)=(dy)/(dt)4-2t,V_(x)=(dx)/(dt)=3`
`rArr u_(y)=v_(y)underset(t=0)(|)=4,u_(x)=v_(x)underset(t=0)(|)=3`
The angle of projection `:`
`tantheta=(V_(y))/(V_(x))=(4)/(3)rArrtheta=tan^(-1)((4)/(3))Ans.`
Promotional Banner

Topper's Solved these Questions

  • DAILY PRACTICE PROBLEMS

    RESONANCE ENGLISH|Exercise dpp 17|7 Videos

  • DAILY PRACTICE PROBLEMS

    RESONANCE ENGLISH|Exercise dpp 18|7 Videos

  • DAILY PRACTICE PROBLEMS

    RESONANCE ENGLISH|Exercise dpp 15|8 Videos

  • CURRENT ELECTRICITY

    RESONANCE ENGLISH|Exercise Exercise|53 Videos

  • ELASTICITY AND VISCOCITY

    RESONANCE ENGLISH|Exercise Advanced Level Problems|9 Videos

Similar Questions

Explore conceptually related problems

The vertical height of the projectile at the time is given by y=4t-t^(2) and the horizontal distance covered is given by x=3t . What is the angle of projection with the horizontal ?

If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is -

What is the relationship between time of flight T and horizontal range R? (where theta is angle of projection with the horizontal)

The vertical height y and horizontal distance x of a projectile on a certain planet are given by x= (3t) m, y= (4t-6t^(2)) m where t is in seconds, Find the speed of projection (in m/s).

The distance s of a particle in time t is given by s=t^3-6t^2-4t-8 . Its acceleration vanishes at t=

The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

The horizontal range of a projectile is 4 sqrt(3) times its maximum height. Its angle of projection will be

The coordinates of a moving particle at any time t are given by, x = 2t^(3) and y = 3t^(3) . Acceleration of the particle is given by

The position of a particle along X-axis at time t is given by x=2+t-3t^(2) . The displacement and the distance travelled in the interval, t = 0 to t = 1 are respectively

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is