Question

A hot-air balloon is 11.0 m above the ground and rising at a speed of 7.00 m/s. A ball is thrown horizontally from the balloon basket at a speed of 9.00 m/s. Ignore friction and air resistance and find the speed of the ball when it strikes the ground.

Answer 1

Answer:

18.6 m/s

Explanation:

$h$ = Initial height of the balloon = 11 m

$v_{o}$ = initial speed of the ball

$v_{oy}$ = initial vertical speed of the ball = 7 m/s

$v_{ox}$ = initial horizontal speed of the ball = 9 m/s

initial speed of the ball is given as

$v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{9^{2} + 7^{2}} = 11.4 m/s$

$v_{f}$ = final speed of the ball as it strikes the ground

$m$ = mass of the ball

Using conservation of energy

Final kinetic energy before striking the ground = Initial potential energy + Initial kinetic energy

$(0.5) m v_{f}^{2} = (0.5) m v_{o}^{2} + mgh \\(0.5) v_{f}^{2} = (0.5) v_{o}^{2} + gh\\(0.5) v_{f}^{2} = (0.5) (11.4)^{2} + (9.8)(11)\\(0.5) v_{f}^{2} = 172.78\\v_{f} = 18.6 m/s$

Answer 2

Answer:

0.95 second

Explanation:

height, h = 11 m

ux = 9 m/s

uy = 7 m/s

Let it takes time t to strike the ground.

Use second equation of motion

$h = u_{y}t + 0.5 gt^{2}$

- 11 = 7 t - 0.5 x 9.8 t²

-11 = 7t - 4.9t²

4.9t² - 7t - 11 = 0

$t=\frac{-7\pm \sqrt{49+4\times4.9\times11}}{9.8}$

$t=\frac{-7\pm 16.27}{9.8}$

take positive sign

t = 0.95 second

Thus, the time taken to reach the ground is 0.95 second.