Question

A certain juggler usually tosses balls vertically to a height 2.4 yd. To what height must they be tossed if they are to spend twice as much time in the air?

Answer 1

Answer:

The balls will be tossed to a height of 9.6 yd

Explanation:

From vertical equation of motion;

H = ut +1/2(gt²)

where;

U is the initial velocity

H is the vertical height

g is acceleration due to gravity

t is the time of motion

Assuming no initial, U = 0

H = 1/2(gt²) ⇒ $t = \sqrt{\frac{2H}{g} }$

Time spent at the initial throw (t₁) to a height of 2.4 yd:

$t_1 = \sqrt{\frac{2*2.4}{g} }$

If the ball spend twice as much time in air, $t_2 = 2t_1$

$H_2 =\frac{g(2t_1)^2}{2}$ $H_2 = \frac{g(2\sqrt{\frac{2*2.4}{g}})^2}{2} = \frac{g(4)(\frac{2*2.4}{g})}{2} =\frac{g(4)({2*2.4})}{2g} = \frac{(4)({2*2.4})}{2} = 4X2.4 = 9.6 yd$

Therefore, the balls will be tossed to a height of 9.6 yd if they are to spend twice as much time in the air.