the water specific heat will remain at 4.184.
After x seconds, an object will fall

where a is acceleration due to gravity and t is time
so when t=3.3
the distance it will fall is

=53.361m
it will fal 53.361 meters
The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
Learn more about momentum:
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Answer:48.2 Joules
Explanation:
Given
two masses of 0.2 kg and 0.4 kg collide with each other
after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east
Velocity of 0.2 kg mass is


Velocity of 0.4 kg mass


Thus total Kinetic energy 
Kinetic energy=48.2 J