Asked by Mishka
A tennis ball is projected vertically upwards from the top of a 55 m cliff by the sea. Its height from the point of projection, s m, ts later is given by
s = 50t - 5t^2
a. when does the tennis ball hit the sea?
b. find the impact velocity of the ball as it enters the sea.
c. find the mean speed of the ball for its entire time of flight.
s = 50t - 5t^2
a. when does the tennis ball hit the sea?
b. find the impact velocity of the ball as it enters the sea.
c. find the mean speed of the ball for its entire time of flight.
Answers
Answered by
henry2,
V = Vo + g*Tr = 0,
50 + (-10)Tr = 0,
Tr = 5 seconds. = Rise time.
S = 50*5 - 5*5^2 = 125 m. = Max ht. above bldg.
125 + 55 = 180 m. = max ht. above gnd.
h = 0.5g*Tf^2 = 180,
5Tf^2 = 180,
Tf = 6 seconds. = Fall time from max ht. to gnd.
a. T = Tr + Tf = 5 + 6 = 11 seconds.
b. V^2 = Vo^2 + 2g*h = 0 + 20*180 = 3600,
V = 60 m/s.
50 + (-10)Tr = 0,
Tr = 5 seconds. = Rise time.
S = 50*5 - 5*5^2 = 125 m. = Max ht. above bldg.
125 + 55 = 180 m. = max ht. above gnd.
h = 0.5g*Tf^2 = 180,
5Tf^2 = 180,
Tf = 6 seconds. = Fall time from max ht. to gnd.
a. T = Tr + Tf = 5 + 6 = 11 seconds.
b. V^2 = Vo^2 + 2g*h = 0 + 20*180 = 3600,
V = 60 m/s.
Answered by
henry2,
c. (Vo+V)/2 = (50+60)/2 = 55 m/s.
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