Question

A child on a swing is 1m above the ground at the lowest point and 6 m above the ground at the highest point. The horizontal speed of the child at the lowest
point of the swing is approximately.
A) 8 ms -1
B) 12 ms -1
C) 10 ms -1
D) 14 ms-1​

Answer 1

Answer:

{\huge {\mathfrak {Answer :-}}}Answer:−

__________

v = final velocity

u = initial velocity

s = Distance

a = Acceleration

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✨ First case,

v = 0 m/s

u = 10 m/s

s = 20 m.

➡️ v^2 = u^2 + 2as

Or, 0 = 100 + 40a

Or, a = - 100 /40 = - 5/2.

.°. a = - 5/2 m/s^2

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✨Second case,

v = 0 m/s

u = 20 m/s

a = - 5/2 m/s^2

➡️ v^2 = u^2 + 2as

Or, 0 = 400 - 5s

Or, s = 400/5 = 80.

.°. The stopping distance will be 80 m.

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Thanks..

Answer 2

Answer: c - 10ms-1

Explanation:

the highest position is 6m and

The lowest position is 1m.

So the swing comes down a height of [6m-1m]= 5m

And at the lowest point there won't be a potential energy and at the heighest point there won't be a kinetic energy,

So, PE+KE = PE+KE

mgh+1/2mv2 = mgh+1/2mv2

0+1/2mv2=mgh+0

V^2= 2*10*5. (Here 5 is the height difference and m gets cancelled bcz in both the sides of the equation there is m )

V= sqrt of 2*10*5

V= sqrt of 100

V = 10ms-1

Hope u understood... :)