A cannonball is shot (from ground level) with an initial horizontal velocity of 34 m/s and an initial vertical velocity of 26 m/s.

Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 34 m/s and an initial vertical velocity of 26 m/s.

1)What is the initial speed of the cannonball?

2)What is the initial angle θ of the cannonball with respect to the ground?

3)\What is the maximum height the cannonball goes above the ground?

4)How far from where it was shot will the cannonball land?

5)What is the speed of the cannonball 4.2 seconds after it was shot?

6)How high above the ground is the cannonball 4.2 seconds after it is shot?

Answer 1

1) We need Pythagoras to find the vector sum v=sqrt(34^2 + 26^2) = 42.8 m/s

2) We use inverse tan of (26/34) = 37.4 degrees

3) At the top the vertical velocity is zero and initial v = 26 we get 0=26-(9.8)t so t = 2.65 s

we now find ht y= 0 +(26)(2.65) - 0.5(9.8)(2.65^2) = 34.5 meters

4) We know the time up = time down so total time = 2.65 x 2 = 5.3 s

Therefore horizontal distance = 5.3 x 34 = 180.2 m

5) Horizontal Speed is constant at 34 m/s, vertical speed is v=26-9.8(4.2) = 15.2 m/s down

Using Pythagoras and vector sum speed = sqrt(34^2 + 15.2^2) = 37.2 m/s

6) Using y = 26(4.2) -(0.5)(9.8)(4.2^2) we get 22.8 meters