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:

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⭐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

Step-by-step explanation:

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