The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s
Distance = (3*5) + (1/2*1*5^2)= 15+12.5= 27.5m
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds).
Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.
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Answered using calculus.
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is
velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)
The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s.
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.
Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)
60 = 9.8t
60 / 9.8 = t
t = 6.122 s
Hopefully this is the right answer.
Answer:
an object sliding down hill
Explanation:
On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.