Question

A body is thrown upwards with the velocity of 12m/s ,find the maximum hight it reached

Answer 1

Answer:

From Newton: s=ut+(1/2)at^2

Where:

s=distance, u=initial velocity, a=acceleration and t=time

If the interval in question is 1 second the equation simplifies to:

s=u+(1/2)a

Initial velocity (up) is 12ms and acceleration due to gravity (down) is -9.81ms^2.

ergo s= 12–4.905 = 7.095m assuming the throw is made on the surface of the Earth.

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Explanation:

Answer 2

Answer:7.35

Explanation:2as=v²-u²

Where, V=0, u=12, a=-9.8

S=v²-u²/2a

S=0²-12²/2(-9.8)

S=-144/-19.6

S=7.35