- #1
Chocolaty
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I answered the questions and I think i have them right, this is new stuff though so I just want to make sure i understand everything, if someone could confirm it would put a smile on my face like this ->
During a serve, a tennis player hits the ball horizontally with a force of 500 N, without putting a spin on it. The 100g ball leaves the racket at a velocity of 40 m/s
a) calculate the time during which the ball was in contact with the player's racket
So i use the impulse formula:
Force * Time = Mass * delta velocity
500 * t = 0.1(kg) * 40
t = 0.008s
b) During its displacement, the ball slows down due to the effect of air resistance. Assuming that the frictional force of 5 N is constant, calculate the ball's average horizontal deceleration in the first second of its flight.
Here i'll find the difference between the initial acceleration and the one after 1 second:
Force = Mass * Acceleration
500 = 0.1 * a
a = 5000m/s^2
Fnet = Fa - Ff = ma
500 - 5 = 0.1 * a
a = 4950m/s^2
So the horizontal deceleration is -50m/s^2
c) Calculate the ball's horizontal velocity when it hits the ground, on the other side of the net, at a distance of 12m from the server.
(Delta V)^2 = 2ad
V2^2 - 1600 = 2 * -50 * 12
V2 = sqr(400) = 20m/s
Thanks in advance
During a serve, a tennis player hits the ball horizontally with a force of 500 N, without putting a spin on it. The 100g ball leaves the racket at a velocity of 40 m/s
a) calculate the time during which the ball was in contact with the player's racket
So i use the impulse formula:
Force * Time = Mass * delta velocity
500 * t = 0.1(kg) * 40
t = 0.008s
b) During its displacement, the ball slows down due to the effect of air resistance. Assuming that the frictional force of 5 N is constant, calculate the ball's average horizontal deceleration in the first second of its flight.
Here i'll find the difference between the initial acceleration and the one after 1 second:
Force = Mass * Acceleration
500 = 0.1 * a
a = 5000m/s^2
Fnet = Fa - Ff = ma
500 - 5 = 0.1 * a
a = 4950m/s^2
So the horizontal deceleration is -50m/s^2
c) Calculate the ball's horizontal velocity when it hits the ground, on the other side of the net, at a distance of 12m from the server.
(Delta V)^2 = 2ad
V2^2 - 1600 = 2 * -50 * 12
V2 = sqr(400) = 20m/s
Thanks in advance