Question

An object is thrown up with a speed of 50 m/s. its height from the ground after 7s is?

Answer 1

Let us first consider first part of motion i.e. when it reaches the highest point

u = 50 m/s

g = -10 m/s^2(for ease of calculation u can also take it as 9.8 which would be more precise) Note - negative sign comes because the direction of velocity and acceleration due to gravity are opposit

v = 0 (as it comes to rest at topmost point)

Applying

2gh = v^2 - u^2

We get, h = 125 m

Now, applying,

v = u+gt, we get -

t = 5 s

So after 5 seconds , the ball will reach its maximum height.

Now for downward journey, remaining seconds = 7 s - 5 s = 2 s

Now, u = 0( coming down from maximum height, it would be at rest initially)

t = 2 s

g = 10 m/s^2

Applying h = ut + 1/2 gt^2 ,we get-

h = 20 m

So the object has traveled 20 m from the topmost point

Therefore it is 125 - 20 meters i.e. 105 meters from the ground

Answer 2

Use the kinematics formula:

h=ut+(gt^2)/2, where

h=final height (to be obtained)

u=initial velocity (50 m/s)

g= acceleration due to gravity (-9.8 m/s^2)

t=time of flight (7 s)

Put the values in the equation to obtain the value of h.

(h comes out to be about 105 m)