Question

A body is projected vertically upwards with a velocity of 21m/s. Find the maximum height, time taken by it to reach this height, the velocity with which it passes a point at a height of 10m. (Please send me the solution with an attachment)

Answer 1

H=u^2/2g
H= 21×21/20
keep H in the formula v=u+at and get T.
h=10
u=21
use v^2-u^2=2as to get v as u and h are given and a is 10

Answer 2

Let it happen at a height of x meters.

Distance = twice |displacement|

=> H + (H-x) = 2 * x

=> 2H = 3x

=> x = 2/3 H

Now, v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time). We know v is zero at max height and u = 20m/s. Assuming acceleration due to gravity is 10m/s^2, we get

0 = 20 - 10t

=> t = 2

i.e. takes two seconds to reach max height.

Again, S = ut + 1/2at^2 (where S = distance). Substituting, we get

H = (20 * 2 ) - 1/2(10 * 2^2)

=> H = 20

thus x = 2/3H = 40/3

Now we just need to find out how long it takes for an object dropped from rest to fall a distance of H-x = 20 - 40/3 = 20/3

Using the equation S = ut + 1/2at^2, we get t = 2/root3

Thus, 2 seconds to get to max height and 2/root3 secs to fall, giving a total time requirement of 2 + 2/root3