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.
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.
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
The angle = 30 degree
ramp length = 6 m
height = 3m
https://www.mathsisfun.com/quadratic-equation-solver.html
That is in case you should need to solve a quadratic.
To answer part (a) of the question, we need to determine the velocity of the puppy the instant it leaves the ramp. We can use the equation of motion:
v^2 = u^2 + 2as
where v = final velocity, u = initial velocity, a = acceleration, and s = displacement.
Given:
Initial velocity, u = 0 (since the puppy starts from rest)
Acceleration, a = 10 m/s^2
Displacement, s = 6 m
Plugging in the values into the equation, we can solve for v:
v^2 = 0^2 + 2 * 10 * 6
v^2 = 120
v = √120
v ≈ 10.954 m/s
Therefore, the velocity of the puppy the instant it leaves the ramp is approximately 10.954 m/s.
Moving on to part (b) of the question, we need to find the horizontal distance the puppy travels from when it leaves the ramp to when it lands on the ground. Since the rocket turns off the instant the puppy leaves the ramp, there is no horizontal acceleration acting on the puppy during this time. Thus, the horizontal motion can be considered as uniformly motion.
We can use the equation of motion for uniformly accelerated motion:
s = ut + (1/2)at^2
where s = displacement, u = initial velocity, a = acceleration, and t = time.
For the horizontal motion, the acceleration is zero, so the equation simplifies to:
s = ut
Given:
Initial velocity, u = 10.954 m/s (from part a)
To find the time, t, we need to determine the time it takes for the puppy to fall from the height of the ramp to the ground. We can use the equation of motion for vertical motion:
s = ut + (1/2)gt^2
where s = displacement, u = initial velocity, g = acceleration due to gravity, and t = time.
Given:
Displacement, s = height of the ramp (which is not given in the question)
Without the height of the ramp, we cannot determine the exact time it takes for the puppy to fall. Therefore, we cannot find the horizontal distance the puppy travels from when it leaves the ramp to when it lands on the ground.
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