Question

While playing catch with my dog I bounced the ball off the ground at a 30-degree angle. It had a range of 6 meters.

Answer 1

The initial velocity of the ball is 8.2 m/s

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A uniform motion along the horizontal direction, at constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration ($g=9.8 m/s^2$, acceleration of gravity)

The range of a projectile can be derived by the equation of motions along the two directions, and it is found to be:

$d=\frac{v^2 sin 2\theta}{g}$

where

v is the initial velocity of the projectile

$\theta$ is the angle of projection

g is the acceleration of gravity

For the ball in this problem, we have

$\theta=30^{\circ}$

d = 6 m is the range

Solving for v, we find the initial velocity:

$v=\sqrt{\frac{gd}{sin 2\theta}}=\sqrt{\frac{(9.8)(6)}{sin (2\cdot 30^{\circ})}}=8.2 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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