Answer to Question #271593 in Calculus for Daemus

Question

show that when a body is dropped from a height h ft above the ground, its velocity v (neglecting air resistance) when it strikes the ground will be v=√2gh where g=32ft/sec^2.

Answer 1

Solution;

When the body is dropped from a height h ft above the ground,initial velocity of the body ,u=0 and it undergoes a free fall.

By the kinematics equation,the final speed is calculated as;

$v^2=u^2+2gh$

Since u=0;

$v^2=0+2gh$

Therefore;

$v=\sqrt{2gh}$

Where g is the gravitational pull,$g=32ft/s^2$