The value of u so that the ball reaches at point A (g = 10m/s2)

A ball is dropped onto a step at a point P and rebounds with velocity u_@ at an angle of 45^@ with the vertical. The value of u_0 knowing that just before the ball hits the point Q its velocity forms an angle 30^@ with the vertical is: (g=10m//s^2)

Track OABCD (as shown in figure ) is smooth. What minimum speed has to be given to a particle lying at point A, so that it can reach point C? (take g=10 m//s^2 )

A small ball is rolled with speed u from point A along a smooth circular track as shown in Fig. 3E.33. If x = 3R , determine the required speed u so that the ball returns to A, the point of projection. What is the minimum value of x for which the ball can reach the point of projection ?

When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres. (a) What was the initial speed of the ball ? (b) How much time is taken by the ball to reach the highest point ? (g=10 m s^(-2))

A boy throws a ball upward with velocity v_(0)=20m//s . The wind imparts a horizontal acceleration of 4m//s^(2) to the left. The angle theta at which the ball must be thrown so that the ball returns to the boy's hand is : (g=10ms^(-2))

A small block of mass m=2kg is at rest point P of the stationary wedge having frictionless hemispherical track of radius R=4m . The minimum constant acceleration 'a_(0)' ( in m//s^(2) ) is to be given to the wedge in horizontal direction, so that the small block just reaches the point Q as shown in figure. Assume all the contacts are frictionless and g=10m//s^(2) . If a_(0)=2xm//s^(2) . Find the value of x

A ball projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of 6 s and the ball reaches the ground after 12 s . The height of the tower is (g = 10 m//s^2) :

A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back ? (g = 10m//s^2)

A particle is placed at the point "A" of a frictionless track "ABC" as shown in figure.It is gently pushed towards right.The speed of the particle when it reaches the point "B" is: (Take g=10m"/"s^(2)

If a ball is thrown vertically upwards with a velocity of 40m//s , then velocity of the ball after 2s will be (g = 10m//s^(2))