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.

Question

a) Find the velocity of the puppy the instant it leaves the ramp
b) Find the horizontal distance the puppy travels from when it leaves the ramp to when it lands on the ground.

Answer 1

Angle of ramp ??????
All I can say is that to go 6 m from rest at 10m/s^2
d = (1/2)a t^2
6 = 5t^2
t = sqrt (6/5) second
speed along ramp at exit S = a t = 10 sqrt(6/5)
Now we are stuck
Vertical problem
Vi = S sin A where A is the ramp slope angle
Hi = initial height = 6 sin A
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2
for ground solve for h = 0, that t is time in air

horizontal problem
u = S cos A forever = 10 sqrt(6/5) cos A

a) S at A

b) u * time in air

Answer 2

The angle = 30 degree
ramp length = 6 m
height = 3m

Answer 3

Now you tell me :)
Well so
Hi = 6 sin 30 = 3 meters high
u = S cos 30 = .866 S
Vi = S sin 30 = .5 S

Answer 4

https://www.mathsisfun.com/quadratic-equation-solver.html

That is in case you should need to solve a quadratic.