Hello. I received help on this question but (c) was not correct and still need h

Question

Hello. I received help on this question but (c) was not correct and still need help. I have included the answers to the (a) and (b) and some comments on help with (c).

A startled armadillo leaps upward rising 0.542 m in the first 0.200 s.
(a) What is its initial speed as it leaves the ground?
3.69 m/s

(b) What is its speed at the height of 0.542 m?
1.73 m/s

(c) How much higher does it go?
? m

Hint: Constant-acceleration equations, for vertical motion. You know the final velocity (at maximum height), the time to reach maximum height, and the (downward) acceleration, and you want the initial velocity. Next, you know the initial velocity, the time to the tower top, and the acceleration, and you want the height of the tower.

Thank you.

Explanation / Answer

Just using the equation v^2 = u^2 + 2as

v^2=o at max height
a=-9.81

rearanging

u^2=2as

(3.69)^2=2 x 9.81 x s

((3.69)^2)/(2x9.81)= s = 0.693990825
0.693990825-0.542=answer=0.151990825=0.152

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