A cannonball was fired horizontally out to sea with a speed of 2.0x10
2 m/s from a cliff top that was
1.5x10
2 m above sea level. It struck a pirate ship anchored in the bay. Determine:
a. the time the cannonball was in flight
b. the distance between the pirate ship and the cliff base
a. To determine the time the cannonball was in flight, we can use the kinematic equation for horizontal motion:
s = ut + 0.5at^2
Where:
s = distance = 1.5x10^2 m
u = initial velocity = 2.0x10^2 m/s
a = acceleration (in the horizontal direction) = 0
t = time
Plugging in the values:
1.5x10^2 = (2.0x10^2)t + 0.5(0)t^2
1.5x10^2 = 2.0x10^2t
t = 1.5/2 = 0.75 seconds
Therefore, the time the cannonball was in flight was 0.75 seconds.
b. To determine the distance between the pirate ship and the cliff base, we can use the equation of motion:
s = vt
Where:
s = distance
v = velocity of the cannonball in the horizontal direction = 2.0x10^2 m/s
t = time = 0.75 seconds
Plugging in the values:
s = (2.0x10^2)(0.75)
s = 150 meters
Therefore, the distance between the pirate ship and the cliff base was 150 meters.