Previously, I asked the question: A ball is thrown vertically upwards with an in

Question

Previously, I asked the question:

A ball is thrown vertically upwards with an initial velocity of 60m/s. After time t seconds, its height above the ground is given as s=60t-16t2. Find its instantaneous velocity after t seconds. Discuss the specialty about time t=15/8

Someone answered with the following:

s=60t-16t2

ds/dt=60-32t

at t=15/8

ds/dt=0

Several other people had the same answer.

Then two of my peers went on to say that was incorrect and said:

this is how you solve it. Instantaneous velocity = change in velocity / change in time v = v initial + accleration X time x - x initial = v initial X t + 1/2 acceleration X time squared Use of equations:60m/s X15/8 + 1/2 0 X*2

Can you bring some clarity to this?

Explanation / Answer

formula is:

s = ut+0.5at^2

(comparing wtih given formula)

s = 60t - 0.5*32*t^2 (given)

thus u = 60 ft/sec

a= -32 ft/sec^2

v = u -at

=> v = 60 - 32*t answer(a)

at time t = 15/8

v = 0 answer(b)

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