A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds after being launched is s(t) = −16t 2 + 32t+ 240 feet above the ground. What is the velocity of the ball as it hits the ground?

V = (-32t +32) ft/s

The velocity of the bowling ball after t seconds is the time derivative of the position function s(t). To obtain the velocity from s(t) we simply differentiate s(t) with respect to t. That is

V = ds(t)/dt = -16×2t + 32×1 = -32t + 32.

When differentiating, you multiply the coefficient of each term in the equation with the power of the variable and then reduce the power by 1 just like above.

Explanation:

you cant move and you feel like someones pushing you down onto your bed it can feel like you want to get up but you just can't it can be extremely scary

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