How Do You Calculate Initial Speed in Projectile Motion Problems?

[heeling23]

1. A ball is thrown horizontally from a height of 22 m and hits the ground with a speed that is three times its initial speed. What was the initial speed?

2. A projectile is fired with initial speed v at an angle 30° above the horizontal from a height of 41 m above the ground. The projectile strikes the ground with a speed of 1.4v. Find v.

These are essentially the same problems right? Except for the fact that #2 has a vertical initial velocity. I tried using V^2=Vo^2+2ah but I got the answer wrong. Then I tried finding the time for the first one using y=1/2at^2 and then using v=vo+at and I got the wrong answer again.

Any help is appreciated!
Thanks,
heeling23

 

[Doc Al]

I tried using V^2=Vo^2+2ah but I got the answer wrong. Then I tried finding the time for the first one using y=1/2at^2 and then using v=vo+at and I got the wrong answer again.

Either approach, done correctly, would work. Post the details of what you did for more help. (Hint: Only the vertical component of motion is accelerated.)

 

[quicknote]

Hello heeling23,

Yes, these two problems are essentially the same, except for the initial vertical velocity in the second problem. In order to solve these problems, you will need to use the equations of motion for projectile motion. These equations are:

1. Horizontal position: x = x0 + v0x * t
2. Vertical position: y = y0 + v0y * t - 1/2 * g * t^2
3. Horizontal velocity: v = v0x
4. Vertical velocity: v = v0y - g * t

In the first problem, the ball is thrown horizontally, so the initial vertical velocity (v0y) is equal to 0. Therefore, the equations become:

1. x = x0 + v0x * t
2. y = y0 - 1/2 * g * t^2
3. v = v0x
4. v = -g * t

We know that the initial height (y0) is 22 m and the final height (y) is 0 m. We also know that the final velocity (v) is 3 times the initial velocity (v0x). So we can set up the following equations:

1. 0 = 22 + v0x * t
2. 0 = 0 - 1/2 * 9.8 * t^2
3. 3v0x = -9.8 * t

We have three equations and three unknowns (v0x, t, and v). We can solve for t in the second equation and substitute it into the first and third equations. This will give us two equations with two unknowns, which we can solve for. The final velocity (v) will be the same as v0x, since there is no change in the vertical velocity. This will give us the initial speed in the first problem.

In the second problem, the initial vertical velocity (v0y) is not equal to 0. So the equations become:

1. x = x0 + v0x * t
2. y = y0 + v0y * t - 1/2 * g * t^2
3. v = v0x
4. v = v0y - g * t

Again, we know the initial height (y0) and the final height (y