Question

An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial velocity of 15 m/s.
-how long was the ball in the air for?
-how far did the ball travel?
-what was the max height of the cannonball?

Answer 1

Answer:

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

$(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]$

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

$y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]$

$0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]$

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

$x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]$

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

$y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]$