Quick Question - Motion? (1 Viewer)

[Zokunu]

A particle is projected vertically upwards from a point 0 with an velocity of 25ms-1 and a downward acceleration of 10ms-1.

a) Find it's velocity and height at any time. V=25 - 10r and X=25t - 5t^2
b) What maximimum height does the particle reach? 31.25m
c) At what time has it's velocity been reduced to half the velocity of projection. 1.25s....How? I got 2.5...

This is a simple one, but i still got it wrong. A body falls from rest. Its velocity v at any time t is given by v = 49(1 - e^-0.5t ). Find

a) It's acceleration at any time?

StepbyStep if possible. Thank you so much.

 

[awesome-0_4000]

c) Using V=25-10r, substitute 12.5 (half of the velocity of projection) for V
12.5=25-10r
10r=25-12.5=12.5
r=1.25

a) Remember, acceleration is the derivative of velocity with respect to time so:
v=49 -49e^-0.5t
dv/dt=24.5e^-0.5t