Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/ sec2-, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?​

Answer 1

Explanation:

initial velocity=40m/s

acceleration due to gravity=10m/S2

maximum height reached by the stone=U/g=40/10=4m

net displacement=0 as the object is thrown upward

total distance covered by the stone=8m

Answer 2

Answer:

The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Explanation:

Given that

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

${v}^{2} = {u}^{2} + 2gh$

$0 - {40}^{2} = 2 \times 10 \times h$

The maximum height is

h=80m

The stone will reach at the top and will come down

The stone will reach at the top and will come down Therefore,the total distance will

$s = h_{1} + h_{2}$

$s = 80m + 80m = 160m$

The net displacement is

$D = h_{1} - h_{2} \\ D = 80 - 80 = 0$

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

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