My wonderful physics problem assistant Burt is standing on a table 2 meters high

Question

My wonderful physics problem assistant Burt is standing on a table 2 meters high. Hekicks a ball directly upward with an initial velocity of 15 m/s. The initial coordinatesof the ball are xi= 0 and yi = 2.

a.) What is the maximum height of the ball relative to the ground?

b.) How long does it take the ball to reach ¾ of its maximum height? I want you todo two things: Find the time to reach its maximum height and then multiply it by ¾. Thenuse a kinematic equation to find the time. Are they the same? Whatever your answer,

please explain why or why not.

Give the answer correct in detail.

Explanation / Answer

given that

initial velocity u = 15 m/s

initial height of ball si = 2 m

part(a)

we know that at the maximum height the velocity (v) is zero.

according to third law of motion

u^2 = v^2 + 2*a*s

where a= -g = -9.8m/s^2

(15)^2 = 0 + 2 * (-9.8)*s

225/19.6 = s

s = 11.47 m

the maximum height of the ball relative to the ground

S = si + s

S = 2 + 11.47

S = 13.47 m

part(b)

if we take maximum height relative to the table

So

H = s*3/4 = 11.47*3/4

H = 8.60 m

the time to reach this height
H = u*t +1/2*a*t^2

8.60 = 15*t - 4.9*t^2

4.9*t^2 - 15*t +8.60 = 0

t1 = 2.29 s

t2 = 0.76 s

ball is passing through the given height two times .first when going upward and second when coming downward.

the time to going upward is less than the time coming downward .

the ball take the time to reach 3/4 of its maximum height is

t' = 0.76 s

the time to reach its maximum height

v = u+a*t

0 = 15 - 9.8*t

t = 15/9.8

t = 1.53 s

so the time to reach 3/4 of maximum height is

t'' = t*3/4

t'' = 1.14 s

so both the time t' and t'' are not same .

because acceleration is present here so there is no direct relationship b/w time and height.

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