5. A beach ball, moving with a speed of 1.27 m/s, rolls off a pier and hits the water 0.75 m from the end of the pier.
a. How high is the pier above the water?
b. If the same beach ball moved off at 4.50 m/s, how far out from the end of the pier would it land?

A. 1.71 m

B. 2.66 m

Explanation:

A. Determination of the height of the pier.

We'll begin by calculating the time taken for the ball to get to the water

This can be obtained as follow:

Horizontal velocity (u) = 1.27 m/s,

Horizontal distance (s) = 0.75 m

Time (t) =?

s = ut

0.75 = 1.27 × t

Divide both side by 1.27

t = 0.75 / 1.27

t = 0.59 s

Finally, we shall determine the height of the pier as follow:

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 0.59 s

Height of pier (h) =?

h = ½gt²

h = ½ × 9.8 × 0.59²

h = 4.9 × 0.3481

h = 1.71 m

Thus, the height of the pier is 1.71 m

B. Determination of the horizontal distance.

Horizontal velocity (u) = 4.50 m/s

Time (t) = 0.59 s

Horizontal distance (s) =?

s = ut

s = 4.5 × 0.59

s = 2.66 m

Thus, if the ball moved at a velocity of 4.50 m/s off the pier, it will land at a distance of 2.66 m from the end of the pier.

multiply 2kg by latent heat of vap