Asked by Cece
A girl throws a ball vertically upward from a building at an initial height of 55.0 m above the ground with an initial speed of 5.00 m/s.
a) What is the velocity of the ball at the highest point in its trajectory?
b) Calculate the velocity of the ball just before it hits the ground.
c) Calculate the total time of flight of the ball.
a) What is the velocity of the ball at the highest point in its trajectory?
b) Calculate the velocity of the ball just before it hits the ground.
c) Calculate the total time of flight of the ball.
Answers
Answered by
bobpursley
a. velocity at highest is zero
b. Initial PE + intial KE = final KE
m*9.81*55+1/2 m 5^2= 1/2 m vf^2
vf^2= 25+2*9.81*55 solve for vf
c. time of flight:
hf=hi+vi*t-1/2 9.8 t^2
0=55+5t-4.9 t^2 solve for t by quadratic equation.
b. Initial PE + intial KE = final KE
m*9.81*55+1/2 m 5^2= 1/2 m vf^2
vf^2= 25+2*9.81*55 solve for vf
c. time of flight:
hf=hi+vi*t-1/2 9.8 t^2
0=55+5t-4.9 t^2 solve for t by quadratic equation.
Answered by
Damon
a) LOL - ZERO
b)
well doing C first (t at ground where h = 0)
h = Hi + vi t - 4.9 t^2
0 = 55 + 5 t - 4.9 t^2
4.9 t^2 - 5 t - 55 = 0
solve quadratic for t when height is zero
B) we know t at ground from part c above
v = Vi - g t
v = 5 - 9.81 t
b)
well doing C first (t at ground where h = 0)
h = Hi + vi t - 4.9 t^2
0 = 55 + 5 t - 4.9 t^2
4.9 t^2 - 5 t - 55 = 0
solve quadratic for t when height is zero
B) we know t at ground from part c above
v = Vi - g t
v = 5 - 9.81 t
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