Question

This puppy is in a rocket. While the rocket is on the ramp it accelerates the puppy with an acceleration of 10m/s2 along the ramp. The puppy starts form rest at the bottom of the ramp and travels 6m along the ramp. The instant the puppy leaves the ramp the rocket turns off. Find a) the velocity of the puppy the instant it leaves the ramp, and b) the horizontal distance the puppy travels from when it leaves the ramp to when it lands on the ground.


....................(60 degrees)
......................../|
....................6 / | 3
....................../_|
.......................?
.....(30degrees) (90 degrees)

Answer 1

For the first part a) :- { Refer to the attachment }

the velocity of the puppy at the instant it leaves the ramp.

acceleration , a = 10 m/s²

at start velocity , u = 0

s = 6 m.

from the equation ,

v² = u² +2as

v² = 0 + 2 × 10 × 6

v² = 120

v = √120 m/sec.

or , v = 10.95 m/sec

For the second part b) :-

The height of the ramp must be required, so your data is quite incomplete.

After leaving the ramp the puppy will do projectile motion at some angle. So for finding the angle the maximum height must be given.

See in these three values "Angle, Range and Height" , one of them must given to calculate other Two.

Hope it Helps :-)