Question

An object is shot upwards, from the ground, with an initial velocity of 120 m/s. How high
will it be after 4.0s?

Answer 1

Answer:

• Distance covered by the object after 4 sec = 401.6 metres

Explanation:

Given

[The Object is shot upwards, therefore g will be negative]

• Acceleration due to gravity, a = g = -9.8 m/s²
• Initial velocity of object, u = 120 m/s
• Time for which it travelled upward, t = 4.0 sec

To find

• Height/distance covered by stone, s =?

Formula required

• The Second equation of motion

        s = u t + 1/2 a t²

[ Where s is distance covered, u is initial velocity, t is time taken and a is acceleration ]

Calculation:

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = ( 120 ) · ( 4 ) + 1/2 · ( -9.8 ) ( 4 )²

→ s = 480 - 78.4  

→ s = 401.6 m

Therefore,

• Object will cover a distance of 401.6 metres after 4.0 seconds.

Answer 2

Given :

• the ground, with an initial velocity of 120 m/s.

• Time = 4 second

To Find :

• Find the height

Solution :

   

$: \implies \: \: \: \: \: \: \: \boxed{ \sf \: s = ut + \frac{1}{2} {at}^{2} }$

Substitute all values :

$: \implies \: \: \: \: \: \: \: \sf \: s =120\times 4 \: + \frac{1}{ \cancel{2}} \times { \cancel{- 9.8} \: \times 4}^{2} \\ \\ \\ : \implies \: \: \: \: \: \: \: \sf \: s = 480 - 4.9 \times 16 \\ \\ \\ : \implies \: \: \: \: \: \: \: \sf \: s = 401.6 \: m$

Hence Object will cover a distance of 401.6 metres