A ball is gently dropped from a height of 30m. If its velocity increase uniform at the rate of 10m/s with what velocity will it strikes the ground ? After what time will it strike the ground?

Explanation:

We know that,

So, by using the third equation of motion, we will find out the final velocity of the ball.

Therefore,

∴ Hence, the required final velocity of the ball with which it strikes the ground is 24.5 m/s. Since, we know that the initial velocity of the ball is 0 m/s and the acceleration of the ball is 10 m/s². So, by using the first equation of motion, we will find out the time of the ball after which it stikes the ground.

Therefore,

∴ Hence, the required time of the ball after which it strikes the ground is 2.45 seconds.

it would be c. predictability

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft