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~christina~
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[SOLVED] projectile basketball
A 40kg basketball player executes a impressive shot. The play-by-play commentator describing the live action says that the student launched herself at an angle of 35 degrees releasing the ball when she was a horizontal distance of 4.31m from the center of the basket.
a) if the ball was released 1.83m above the floor and the basket is a height of 3.05m above the floor, what was the velocity of the ball when it was released?
b)what was the maximum height of the ball above the floor?
c) what is the velocity of the ball as it enters the basket?
d) What was teh time of flight of the ball?
e) how does the magnitude of the force that the player uses to accelerate the basketball compare (that is less than, equal to, or greater than) with the magnitude of the force that the ball exert on the player?
Use correct physics principles to explain your reasoning..
Vxf= vxi + axt
Xf= xi + vxi*t + 1/2 axt^2
Well I know that the
angle = 35 deg
sx= 4.31m
sy= 3.05m- 1.83m= 1.22m
Vo= ?
t= ?
I need Vo so I plugged in...
Sx= Sox +vx*t + 0.5axt^2 (Sox= 0 and no acceleration in x direction a= 0)
Sx= vx*t
Vx= vo cos theta
Sx= vo (cos theta)*t
Sy= Soy+ Voy *t + 0.5gt^2 (Soy= 0 g= -9.8)
Voy= Vo sin theta
Sy= Vo (sin theta) *t - 4.9t^2
since I need the initial velocity...
I think I plug into the other equation the equation for the distance in the y direcction ...
Sx= vo *cos theta *t
t= Sx/ (vo*cos theta)
Sy= vo sin theta*t - 4.9 t^2
Sy= vo sin theta (Sx/ vo cos theta)- 4.9 (Sx/ vo cos theta)^2
Sy= (tan theta *Sx) - 4.9 (Sx/vo cos theta)^2
Sy- tan theta* Sox = -4.9 ( Sx/ vo cos theta) ^2
[tex]\sqrt{} (Sy- tan theta*Sox) / -4.9[/tex]= Sx/ vo cos theta
Vo cos theta = Sx/ ([tex]\sqrt{} (Sy- tan theta*Sox) / -4.9[/tex])
Vo = need
theta= 35 deg (assuming same angle when player jumps)
Sx= 4.31m
Sy= 1.22m (after subtracting the different heights)
Vo cos 35 = 4.31m / ([tex]\sqrt{} (1.22m- tan 35*4.31m) / -4.9[/tex])
Vo cos 35= 7.12
Vo= 8.69m/s^2
Well I think I did my math correctly...could someone check it for me?
b.)
Max height of the ball...
Vy= 0 at max height...
time at max height is = ?
Well I think ..
Vy= Vo sin theta - gt
0= Vo sin theta - gt
t= Vo sin theta/ g (since negatives cancel out)
t= (8.69m/s^2) (sin 35)/ 9.8
Then t= 0.5086s to reach max height
I don't know if this is fine but...
plugging that into y distance equation..
Sy= Soy+ Vo sin theta*t - 4.9 t*2
Soy= 0
Vo= 8.69 m/s^2
t= .5086s
Sy= 4.98(0.5086s)- 4.9 (0.5086s)
Sy= 1.265m (max height if only including the part that I calculated after subtracting the difference in the original height)
Sy actual= 1.265 + ?= ( I really don't know what I add [can't figure it out])
c.) Assuming the previous parts were okay..( I need help with the above to find out what to add)
velocity of the ball as it enters basket...
I'm not sure how to find his...
Can someone check and help me out...?
Thanks Very much
Homework Statement
A 40kg basketball player executes a impressive shot. The play-by-play commentator describing the live action says that the student launched herself at an angle of 35 degrees releasing the ball when she was a horizontal distance of 4.31m from the center of the basket.
a) if the ball was released 1.83m above the floor and the basket is a height of 3.05m above the floor, what was the velocity of the ball when it was released?
b)what was the maximum height of the ball above the floor?
c) what is the velocity of the ball as it enters the basket?
d) What was teh time of flight of the ball?
e) how does the magnitude of the force that the player uses to accelerate the basketball compare (that is less than, equal to, or greater than) with the magnitude of the force that the ball exert on the player?
Use correct physics principles to explain your reasoning..
Homework Equations
Vxf= vxi + axt
Xf= xi + vxi*t + 1/2 axt^2
The Attempt at a Solution
Well I know that the
angle = 35 deg
sx= 4.31m
sy= 3.05m- 1.83m= 1.22m
Vo= ?
t= ?
I need Vo so I plugged in...
Sx= Sox +vx*t + 0.5axt^2 (Sox= 0 and no acceleration in x direction a= 0)
Sx= vx*t
Vx= vo cos theta
Sx= vo (cos theta)*t
Sy= Soy+ Voy *t + 0.5gt^2 (Soy= 0 g= -9.8)
Voy= Vo sin theta
Sy= Vo (sin theta) *t - 4.9t^2
since I need the initial velocity...
I think I plug into the other equation the equation for the distance in the y direcction ...
Sx= vo *cos theta *t
t= Sx/ (vo*cos theta)
Sy= vo sin theta*t - 4.9 t^2
Sy= vo sin theta (Sx/ vo cos theta)- 4.9 (Sx/ vo cos theta)^2
Sy= (tan theta *Sx) - 4.9 (Sx/vo cos theta)^2
Sy- tan theta* Sox = -4.9 ( Sx/ vo cos theta) ^2
[tex]\sqrt{} (Sy- tan theta*Sox) / -4.9[/tex]= Sx/ vo cos theta
Vo cos theta = Sx/ ([tex]\sqrt{} (Sy- tan theta*Sox) / -4.9[/tex])
Vo = need
theta= 35 deg (assuming same angle when player jumps)
Sx= 4.31m
Sy= 1.22m (after subtracting the different heights)
Vo cos 35 = 4.31m / ([tex]\sqrt{} (1.22m- tan 35*4.31m) / -4.9[/tex])
Vo cos 35= 7.12
Vo= 8.69m/s^2
Well I think I did my math correctly...could someone check it for me?
b.)
Max height of the ball...
Vy= 0 at max height...
time at max height is = ?
Well I think ..
Vy= Vo sin theta - gt
0= Vo sin theta - gt
t= Vo sin theta/ g (since negatives cancel out)
t= (8.69m/s^2) (sin 35)/ 9.8
Then t= 0.5086s to reach max height
I don't know if this is fine but...
plugging that into y distance equation..
Sy= Soy+ Vo sin theta*t - 4.9 t*2
Soy= 0
Vo= 8.69 m/s^2
t= .5086s
Sy= 4.98(0.5086s)- 4.9 (0.5086s)
Sy= 1.265m (max height if only including the part that I calculated after subtracting the difference in the original height)
Sy actual= 1.265 + ?= ( I really don't know what I add [can't figure it out])
c.) Assuming the previous parts were okay..( I need help with the above to find out what to add)
velocity of the ball as it enters basket...
I'm not sure how to find his...
Can someone check and help me out...?
Thanks Very much