Question

can you guys do the last question the bonus question

A cannonball is shot from a cannon, at an initial height of 1.1 meters. At the peak of the trajectory, it strikes a smiley face picture, located 4.1 meters to the right. (define the i-axis to the right, and define the j-axis as up) Hint: In order to solve this, you have to understand what ''At the peak of the trajectory'' means. It means at the maximum height of the cannonball. The initial speed of the cannonball is 9.0 m/s, and it is shot at an angle of 40 degree. (a) Calculate the initial velocity (Remember: Velocity is a vector!) (b) Calculate the velocity at the peak of the trajectory, so just before it strikes the picture. Calculate the time in the air before it strikes the picture. Bonus: You shoot a second cannonball, but it misses the picture, so it lands on the ground. How long is it in the air?

Answer 1

You do not need the mass of the ball.........Neglect air resistance

PE = m * g * height

KE = 1/2 * m * v^2

By conservation of energy:

m * g * height = 1/2 * m * v^2

divide both sides by m/2

2 * g * height = v^2

2 * 9.81 * 1.1 = v^2

v = 4.6 m/s

The component vertically is = v sin 45 =4.6 sin 45=3.91m/s

So.......

2(9.18) h = (4.6 sin 45)^2

h = (15.2)^2 /2(9.18)

h=.1.5 m...........

Answer 2