Question

A stone in fired vertically upwards and lands 5s later, what ws the initial velocity of the stone? for how long was the stone at a height of 20m or more?

Answer 1

initial velocty i found= 24.525ms

Answer 2

@Ishaan94 @SmoothMath @FoolAroundMath @ujjwal

Answer 3

Use the relation: \(S=ut-\frac{1}{2}gt^2\) and put S=0, the value of u obtained will be initial velocity.

Answer 4

i found that already

Answer 5

The time taken by stone to reach max height is 2.5 sec. Find the time taken by stone to reach height 20 m. let it be t. Now the time for which stone will be above 20m is 2*(2.5-t)

Answer 6

for the duration of 5 seconds, total displacement is zero (because the ball lands back at the ground).
So, using \(s = ut + \frac{1}{2}at^{2}\), and setting s = 0, you ca get the initial velocity which has been correctly found by you.

Now, we need to find the time at which the stone was at a height of 20m. We will use \(s = ut + \frac{1}{2}at^{2}\) and as it is a quadratic, we will get two solutions for \(t\). The first time is when the ball is moving upwards and first reaches the height of 20m. The second time is when the ball is moving downwards and is at a height of 20m.

The difference between these two times will give you the duration for which the ball was at a height of greater than 20m

Answer 7

For the second part, use FoolAroundMath's process..

Answer 8

acceleration = -9.8 m/s^2
velocity = -9.8t + C
Height = -4.9t^2 +Ct +K

h(0) = 0 so
-4.9(0)^2 + C(0) +K = 0 so
K = 0
height = -4.9t^2 + Ct

h(5) = 0 so
-4.9(5)^2 +C(5) = 0
C = 24.5 so
height = -4.9t^2 + 24.5t
and
velocity = -9.8t + 24.5

Initial velocity:
v(0) = -9.8(0) +24.5 = 24.5

Now, setting height equal to 20 gives us an easily solvable quadratic which should yield two answers.
h(t) = 20
-4.9t^2 +24.5t = 20
-4.9t^2 + 24.5 - 20 = 0
Using quadratic formula:
t =1.0274622765651211
t=3.9725377234348787

Answer 9

which one to take? t=1 or t= 3.97

Answer 10

Their difference.
3.97-1.02

Answer 11

Didn't you read FoolAroundMath's answer completely? Its clearly mentioned at the end.

Answer 12

Zaphod, we have an equation for height, right? Setting that equation equal to 20 tells us WHEN the ball is at the height of 20 meters. Since it's a quadratic, we get two times. That's GOOD! The ball is at that height twice! Once going up, and once going down.


What does the question ask us? How long was it at 20 or higher.

Answer 13

t1=1.02
t2=3.97

So, object remains above 20 m for (3.97-1.02) sec..

Excellent drawing @SmoothMath !!