A DIVER JUMPS UP OFF A PIER AT AN ANGLE OF 45 DEGREES WITH AN INITIAL VELOCITY OF 4.5 M/S. HOW LONG DOES IT TAKE TO HIT THE WATER? AND HOW FAR FROM THE PIER WILL THE DIVER HIT THE WATER? (ASSUME THE LEVEL OF WATER IS THE SAME AS THE PIER)

0.65 seconds

2.07m

Explanation:

Here we need to look at the vertical and the horizontal components of the initial velocity.

In terms of his hitting the water.

Define the upwards direction as positive

We know his initial upwards velocity (u) is 4.5sin(45)

His displacement (s) when he hits the water is zero

Acceleration due to gravity is -9.8

We want to find t

So using s=ut+1/2a

0=4.5sin(45)t+1/2(-9.8)

Take out a common t

0=t(4.5sin45-4.9t)

t=0   or t= 0.65

The first answer is not valid so the answer is 0.65.

Now we can look at the horizontal component

It takes him 0.65 seconds to hit the water, he travels at a constant horizontal velocity of 4.5cos45

Therefore using s=vt

s=4.5cos45 x 0.65= 2.07m away

the answer is to check all of them! they all apply well at least i think so!

0.19 m/s and 0.45 m/s