Home
Class 11
PHYSICS
Two projectiles are projected at angles ...

Two projectiles are projected at angles `(theta)` and `((pi)/(2)-theta)` to the horizontal respectively with same speed 20m/sec. One of them rises 10m heigher than the other. Find the angles of projection. (Take `g=10m//s^(2)`)

Text Solution

Verified by Experts

Maximum height H = `(u^(2)sin ^(2) theta)/( 2 g) rArr h_1 = ((20)^(2) sin ^(2) theta)/(2g) = 20 sin^(2) theta & h_2 = ((20)^(2) sin^(2) (pi//2- theta))/(2g) = 20 cos^(2) theta`
`h_2 - h_1 = 20 [ cos^(2)theta - sin ^(2) theta] = 10 rArr 20 cos 2 theta = 10 rArr cos2 theta = (1)/(2) rArr 2 theta = 60^(@) rArr theta = 30^(@) and theta' = 90^(@) - theta = 60^(@)`
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN |Exercise BEGINNER S BOX-1|6 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN |Exercise BEGINNER S BOX-2|7 Videos

  • KINEMATICS

    ALLEN |Exercise EXERCISE-2|89 Videos

  • MISCELLANEOUS

    ALLEN |Exercise Question|1 Videos

Similar Questions

Explore conceptually related problems

Two projectiles are projected at angles ( theta) and ((pi)/(2) - theta ) to the horizontal respectively with same speed 20 m//s . One of them rises 10 m higher than the other. Find the angles of projection. (Take g= 10 m//s^(2) )

A ball is thown at angle theta and another ball is thrown at angle (90^(@)- theta) with the horizontal direction from the same point each with speeds of 40 m/s. The second ball reaches 50 m higher than the first ball. Find the individual heights. g= 10 m//s^(2)

A particle of mass m is projected with speed u at an angle theta with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

A body is projected vertically up. What is the distance covered in its last second of upward motion? (g=10 m//s^(2))

A ball is projected with speed u at an angle theta to the horizontal. The range R of the projectile is given by R=(u^(2) sin 2theta)/(g) for which value of theta will the range be maximum for a given speed of projection? (Here g= constant)

In the diamram shown in figure , match the following (g=10m//s^(2))

A particle is projected with a speed v and an angle theta to the horizontal. After a time t, the magnitude of the instantaneous velocity is equal to the magnitude of the average velocity from 0 to t. Find t.

A ball is projected vertically up wards with a velocity of 100 m/s. Find the speed of the ball at half the maximum height. (g=10 m//s^(2))

A particle is projected up the inclined such that its component of velocity along the incline is 10 m//s . Time of flight is 2 sec and maximum height above the incline is 5m . Then velocity of projection will be

A particle is projected with a velocity 10 m/s at an angle 37^(@) to the horizontal. Find the location at which the particle is at a height 1 m from point of projection.