# Initial velocity of an object colliding with another

1. Nov 2, 2015

### JulienB

Hi! I just started studying physics in German, and I am stuck about basic formulas partially because of the language.

1. The problem statement, all variables and given/known data

A ball is launched from the ground at a speed |vb0| = 90m/s with an angle of β = 50° to the horizontal axis x. At a distance of d = 60m is placed a canon, which shoots up perpendicularly to the horizontal axis at the same time as the ball is launched. With what initial velocity |vk0| must the bullet be shot from the canon in order to meet the ball?
Both the ball and the bullet are point-shaped and the air resistance is to be ignored. The relevant equations must be used in their vectorial form.

2. Relevant equations

Anything related to initial velocity and displacement from early courses of mechanics.

3. The attempt at a solution

It looks so simple that I feel ashamed to post here... I started by trying to find at what height yh the ball will be after the distance x = d, and for that purpose I used a formula from Wikipedia:
yh = y0 + d⋅tan β - (g⋅d2)/(2(v0⋅cosβ)2)
= 0 + 60⋅tan50° - (9.81⋅602)/(2(90⋅cos50°)2) = 66.23 m

Now I know the bullet is launched at the same time, but I still don't know how long it took for the ball to reach the height vh. Maybe t = d/(v0⋅cosβ) = 1.037s? That seems very short to reach 66.23m high!
I am also unsure on what formula to use to calculate the initial velocity of a vertical movement. Would someone be so kind to indicate me what formulas should be used in such a problem?

J.

2. Nov 2, 2015

### Staff: Mentor

Hi JulienB.

Your time and height for the projectile look okay to me.

For the vertical projectile, write another projectile equation of motion y(t) = ? . This time there's no x-component to deal with.

3. Nov 2, 2015

### JulienB

Thank you very much for your answer. Would that work with the equation y = v0 ⋅ t ⋅sin 90 - (1/2)g ⋅ t2? Using this, I obtain a plausible initial velocity of 68.78 m/s.

4. Nov 2, 2015

### Staff: Mentor

Sure, that works. Or you could just write directly $y = v_o t - \frac{1}{2} g t^2$. The launch angle of 90° is implied if it is vertical, and you have 1-dimensional motion with acceleration.