The maximum height attained by a ball projected with speed ` 20 ms^(-1)` at an angle `45^(@)` with the horizontal is [take g = ` 10 ms^(-2)`]

A particle is projected with a speed 10sqrt(2) ms^(-1) and at an angle 45^(@) with the horizontal. The rate of change of speed with respect to time at t = 1 s is (g = 10 ms^(-2))

A particle is projected with a speed 10sqrt(2) ms^(-1) and at an angle 45^(@) with the horizontal. The rate of change of speed with respect to time at t = 1 s is (g = 10 ms^(-2))

A body is projected with a velocity 20 m*s^(-1) , making an angle of 45^@ with the horizontal. Calculate the maximum height it can attain

A ball is thrown with a speed of 20 m s^(-1) at an elevation angle 45^(@) . Find its time of flight and the horizontal range ( take g = 10 ms ^(-2)] lt

A ball is thrown with a speed of 20 m s^(-1) at an elevation angle 45^(@) . Find its time of flight and the horizontal range ( take g = 10 ms ^(-2) )

A body is projected at an angle of 30^@ with the horizontal and with a speed of 30 ms^-1 . What is the angle with the horizontal after 1.5 s ? (g = 10 ms^-2) .

A body is projected at an angle of 30^@ with the horizontal and with a speed of 30 ms^-1 . What is the angle with the horizontal after 1.5 s ? (g = 10 ms^-2) .

A ball is projected from ground with a speed of 20ms^(-1) at an angle of 45^(@) with horizontal.There is a wall of 25m height at a distance of 10m from the projection point. The ball will hit the wall at a height of

What are the angles of projection of a projectile projected with velocity 30 ms^(-1) , so that the horizontal range is 45 m ? Take g = 10 ms^(-2) .

A ball is projected with a velocity 20 sqrt(3) ms^(-1) at angle 60^(@) to the horizontal. The time interval after which the velocity vector will make an angle 30^(@) to the horizontal is (Take, g = 10 ms^(-2))