Question

1.) A ball is thrown straight up with an initial speed of 15.715.7 m/s. At what height above its initial position will the ball have one‑half its initial speed?

2.) A child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a speed of 1.101.10 m/s. The frictional force acting on the sled is one‑fifth of the combined weight of the child and the sled. If she travels for a distance of 27.527.5 m and her speed at the bottom is 4.054.05 m/s, calculate the angle that the slope makes with the horizontal.

Answer 1

Part (a)

Given :

V_i =15.7 m/s

V_f =V_i/2

Let S be the hight where the velocity become half of initial velocity.

Now by using the equation of motion:

As the direction of velocity is upward where as the direction of acceleration due to gravity is downward so,

a = - g = - 9.8 m/s²

So,

S = 9.43m

Part(b)

Let,

The child with sled as a block of mass 'm'

Hight of initial point is 'h'

Distance travlled 's'

Normal reaction 'N'

Friction force 'f'

Angle of slope is

Initial velocity V_i

Final velocity V_f

Now, by using work power energy theorem

Work done by all forces = change in kinetic energy

As from above diagram

And by given information

So,

By putting the value in above equation we get

Now from the above diagram

13.19°