Question

46. a)From the top of a tower , a ball is dropped to fall freely under gravity and at

the same time, another ball is thrown up with a velocity of 50m/s. Plot the

position-time graph for the motion of two balls during the time intervals t=0s,

t=2s, t=4s, t=6s, t=8s and t=10s. (g=10m/s2

)

b)The acceleration ‘a’ in ms-2 of a particle is given by a = 5t2 + 3t + 2, where t is

the time. If the particle starts out with a velocity v = 2 ms-1 at t = 0, then find

the velocity at the end of 3s. plzz give a perfect anser

​

Answer 1

Answer:

For the dropped body...

The initial displacement of the body (with respect to ground) be '0' and the final displacement would be 's'.

now,

v = u + at

or initial velocity

u = -at = -10x5 = -50 m/s

and the displacement after 5 seconds will be

s = -u2 / 2a = -2500 / 2x10

or

s = 125m

so, the position-time graph in 5 seconds would be

..

Let the initial displacement be zero.

For the launched body...

u =50 m/s

t = 5s

a = -10 m/s2

so,

displacement after 5 seconds will be

s = ut + (1/2)at2

= 50x5 - (1/2)x10x52

so,

s = 125m