1. assume the acceleration of gravity is 2 10 m/sec downwards. a cannonball is fired at ground level. if the cannon ball rises to a height of 80 meters and travels a distance of 240 meters before it hits the ground, what is the magnitude of the initial veloci ty in meters per second?

A. 50 m/s

Explanation:

So, we know that at the highest point of the cannonball’s path the y-component of the velocity is 0 since the projectile is not moving in that axis momentarily. We can set the following equation at that instant:

Now that we found the y-component of the cannonball’s initial velocity, we can also find how much time it takes the projectile to reach the maximum height point, using the following equation:

We already stablished that in this point vy = 0. In that case:

In the cannonball’s path, the maximum height point is in the middle point of the total x displacement. In other words, the projectile takes about 2t to touch the ground. That means that for the projectile to reach x=240 m, it requires t = 2*4s = 8 s. now, we proceed and calculate the x-component of the initial velocity, using the following equation:

The projectile started at the point of reference, so we consider x0 = 0. Also, the velocity of the projectile maintains constant along the x-axis, so we consider ax=0 as well. Therefore, we have that:

The magnitude of the initial velocity can be found using the Pythagorean Theorem

So, the cannonball must be thrown with an initial velocity of 50 m/s to reach an height of 80 m and a displacement on the x-axis of 240 m

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