I have no idea what equations to use to solve this problems! I have the solution

Question

I have no idea what equations to use to solve this problems! I have the solutions I just need to know how to do them. Thank you :)

Jane intends to retrieve her baseball from the roof of her neighbor's garage. She tries to toss the ball (over a fence) and abck into her own yard. At the instant when the ball leaves Jane's hand, the ball is traveling with a velocity of 6.21 m/s, directed 60.0 degrees above the horizontal. At this instant, the ball is located 0.700 m above the garage roof, at a horizontal distance of 1.5 m from the edge of the garage roof, and at a height of 2.70 m above the level of the fence top. The top of the fence is located at a horizontal distance of 3.00 m from the near wall of the garage.

1) At the instant when the ball has traveled a horizontal distance of 4.5 m from its launch point, the ball is located approximately...

a) 1.3 m below its launch point

b) 2.9 m above its launch point

c) 2.5 m below its launch point

d) 2.5 m above its launch point

e) the ball hits the ground before it reaches the fence

Explanation / Answer

Horizontal velocity is constant, there is no acceleration horizontally since gravity only acts downward

We can find the horizontal velocity from

vx = vcos(angle)

vx = (6.21)(cos60) = 3.105 m/s

Then apply d = vt to find the time required to travel 4.5 m horizontally

4.5 = (3.105)(t)

t = 1.45 sec

That is the amount of time gravity has to act on it, so we can find the distance it travels vertically

Apply d = vot + .5at2

The initial y velocity = vsin(angle) = 6.21(sin 60) = 5.378 m/s

Then d = (5.378)(1.45) + (.5)(-9.8)(1.45)2

d = -2.50 m

Thus the ball is now 2.5 m BELOW the launch point (the negative means below) - That is Choice C

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