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

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

Step-by-step 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)-(1)/(2)*g*t^(2)   \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]`

`0=8.6*t-(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+(1)/(2) *g*(t)^(2) \\y = 0*t+(1)/(2) *9.81*(1.75)^(2)\\ y=15 [m]`