Question

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?

Answer 1

Answer:

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:

$v_y =v_0y^2 +2gh \rightarrow 0 = v_0y^2 +2gh \rightarrow v_0y^2 =2gh$

$v_0y^2 =-2*(-10 m/s^2)*80 m=1600 m^2/s^2$

$v_0y =\sqrt(1600 m^2/s^2) = 40 m/s$

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:

$v_y = v_0_y +a*t$

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

$v_0_y = -g*t \rightarrow 40 m/s=-(-10m/s^2)*t \rightarrow t=(40 m/s)/(10m/s^2) = 4s$

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:

$x = x_0 + v_0x+1/2*a_x*t^2$

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:

$x = v_0x*t \rightarrow 240 m = v_0x*(8s)$

$v_0x = (240 m)/(8s) = 30 m/s$

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

$v_0 = \sqrt{(30 m/s)^2 +(40 m/s)^2}= 50 m/s$

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