Answer all questions.A cannonball is launched horizontally with an initial velocity of 36 m s–1 from the edge of a cliff 200 m high. Assume that air resistance is negligible.(a) Calculate the total time that the cannonball is in the air.[2](b) Determine the horizontal distance from the bottom of the cliff to where the cannonball hits the ground.[2](c) The mass of the cannonball is then doubled. Assume launch speed is the same. Explain whether this will affect the answer in part (b).

(a) The total time that the cannonball is in the air can be calculated using the equation of motion:

h = 1/2gt^2

where: h = height = 200 m g = acceleration due to gravity = 9.8 m/s^2 t = time

Rearranging the equation to solve for t gives:

t = sqrt((2h)/g) = sqrt((2*200)/9.8) = sqrt(40.82) = 6.39 seconds

So, the cannonball is in the air for approximately 6.39 seconds.

(b) The horizontal distance from the bottom of the cliff to where the cannonball hits the ground can be calculated using the equation:

d = vt

where: d = distance v = initial velocity = 36 m/s t = time = 6.39 seconds

Substituting the given values into the equation gives:

d = 36 m/s * 6.39 s = 230.04 m

So, the cannonball hits the ground approximately 230.04 m from the bottom of the cliff.

(c) Doubling the mass of the cannonball will not affect the horizontal distance it travels before hitting the ground. This is because the horizontal motion of the cannonball (its initial velocity) is independent of its mass. The only forces acting on the cannonball after it is launched are gravity (which pulls it downwards) and the initial force of the launch (which propels it forwards). Neither of these forces are affected by the mass of the cannonball, so the horizontal distance it travels remains the same.