Projectile Motion with Unknown Initial Height

[a1234]

Homework Statement

I am given a problem where a 6kg object is launched from a cannon from an unknown initial height and eventually lands on top of a platform that is at a height of 40 meters above the ground. As you can see from the diagram, the final height is greater than the initial height.

The total displacement in the x-direction is 100 meters, and the object is launched from an angle of 25 degrees with an initial velocity of 5 m/s. I am asked to find the initial height at which the object is launched.

Relevant Equations

y=y0+vy0(t)-1/2gt^2,x=x0+vx0(t),v=v0+at, vy^2 = vy0^2-2g*change in y

First, I tried solving for the total time of flight, which I got as 100 = 5cos25*t --> t=22 s
Since we know the height at which the object lands, but not at which it is launched, I tried setting up the equation as:
yf = 40 - y0 = y0 + 5sin25*(22) - 1/2(9.8)(22)^2
However, I got y0 = 1183 m, which is not realistic given the problem statement. I assume this equation works if we only have freefall from an initial height.

I then tried solving for the the height at which vy = 0 (at max height):
0^2 = (5sin25)^2-2*9.8*deltay
delta y = 0.23 m

I also tried vy = v0y - gt for vy = 0 and got t = 0.22 s.

I don't know where to proceed from there. I also don't know if we need to change the sign of acceleration due to gravity when we consider motion past the point of maximum height.

 

[PeroK]

Do you think you can hit a target 100m away by launching a projectile at 5m/s?

 

[a1234]

What would be the approach to solving the problem if the target were not as far away?

 

[PeroK]

What would be the approach to solving the problem if the target were not as far away?

Your answer of over 1000m might well be correct!

 

[PeroK]

What would be the approach to solving the problem if the target were not as far away?

You could try with $v_0 = 50m/s$.

 

[a1234]

With v0 = 50 m/s, I get y0 = 973 m, which still doesn't meet the criterion that y0 < yf.

I have a very similar problem with v0 = 8 m/s, θ = 35 degs, x = 7 m, yf = 3 m, and mass of object = 2 kg. I'm having the same problem with that too.

 

[PeroK]

With v0 = 50 m/s, I get y0 = 973 m, which still doesn't meet the criterion that y0 < yf.

That is definitely not right. What's your time of flight?

With v0 = 50 m/s, I get y0 = 973 m, which still doesn't meet the criterion that y0 < yf.

I have a very similar problem with v0 = 8 m/s, θ = 35 degs, x = 7 m, yf = 3 m, and mass of object = 2 kg. I'm having the same problem with that too.

One problem at a time!

 

[a1234]

Oops...I used the same time of flight as for the previous problem.

The new time is 100 = 50cos25 * t ---> t = 2.2 s
So, 40 - y0 = y0 + 50sin25*2.2 - 1/2(9.8)(2.2)^2 --> y0 = 8.6 m