Home
Class 11
PHYSICS
The equation of motion of a body are gi...

The equation of motion of a body are given such that horizontal displacement x = 3t meter and vertical displacement , `y = 4t- 5t^(2)` meter , where "t" is in seconds. Find the angle of projection , velcoity of projection and horizontal range?

Text Solution

Verified by Experts

The correct Answer is:
`53^(@), 5ms^(-1),2.4m`
Promotional Banner

Similar Questions

Explore conceptually related problems

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 equations of motion of a projectile are given by x=36t m and 2y =96t-9.8t^(2)m . The angle of projection is

The displacement of a particle is given by y=(6t^2+3t+4)m , where t is in seconds. Calculate the instantaneous speed of the particle.

The equations of motion of a projectile are given by x=36tm and 2y =96t-9.8t^(2)m . The angle of projection is

The position coordinates of a projectile projected from ground on a certain planet (with an atmosphere) are given by y = (4t – 2t^(2)) m and x = (3t) metre, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is -

The vertical displacement of a block A in meter is given by y=t^(2)//4 where t is in second. The downward acceleration a_(B) of a block B ( in m//s^(2) ) is

The angular displacement of a particle is given by theta = t^4 +t^3 +t^2 +1 where ‘t’ is time in seconds. Its angular velocity after 2 sec is

The horizontal and vertical displacements x and y of a projectile at a given time t are given by x= 6 "t" and y= 8t -5t^2 meter.The range of the projectile in metre is:

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 ?

The height y nad the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t^2) m and x = 6t m , where t is in seconds. The velocity with which the projectile is projected at t = 0 is.