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A pumpkin is thrown horizontally off of a building at a speed of 2.5 m/s and travels a horizontal distance of 12 m before hitting the ground.
What was the initial height of the pumpkin?
What was the final vertical velocity of the pumpkin?

Respuesta :

xmartha

Answer:

-47.04

Explanation:

Divide 12 m by 2.5 m/s which equals 4.8s

And then use 4.8s to multiply it with -9.8

4.8(-9.8)= -47.04 m/s

Hope this helped!! :)

The initial height of the pumpkin is 23.52m and the final vertical velocity of the pumpkin is 47.04 m/s.

Laws of motion:

Initial horizontal speed, u = 2.5 m/s

horizontal distance covered , d = 12m

So the time taken is:

t = d/u

t = 12/2.5 s

t = 4.8s

Now, the initial vertical speed, v = 0

So the vertical distance traveled by the pumpkin will be equal to the initial height of the pumpkin:

s = vt + ¹/₂gt²

s = 0.5 × 9.8 × 4.8 m

s = 23.52m

Final vertical velocity:

v' = v + gt

v' = 9.8 × 4.8 m/s

v' = 47.04 m/s

Learn more about laws of motion:

https://brainly.com/question/26083484?referrer=searchResults

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