Question

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with a speed of 10.0 m/s. What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground if the hot-air balloon is rising at 6.0 m/s relative to the ground during this throw?

Answer 1

Answer:

Explanation:

Given

balloon is rising with a speed of $u_y=6\ m/s$

Person throws a ball out of basket with a horizontal velocity of $u_x=10\ m/s$

Considering upward direction to be positive

When ball is thrown it has two velocity i.e. in upward direction and in horizontal direction so net velocity is

$v_{net}=\sqrt{(u_x)^2+(u_y)^2}$

$v_{net}=\sqrt{(6)^2+(10)^2}$

$v_{net}=\sqrt{36+100}$

$v_{net}=\sqrt{136}$

$v_{net}=11.66\ m/s$

Direction of velocity

$\tan \theta =\dfrac{u_y}{u_x}$

$\tan \theta =\dfrac{6}{10}$

$\theta =30.96^{\circ}$

where $\theta$ is angle made by net velocity with horizontal .