Projectile Motion of half court basketball shot

[halo9909]

Homework Statement

1)A basketball player tries to make a half-court jump-shot, releasing the ball at the height of the basket. Assuming the ball is launched at 53.0°, 13.5 m from the basket, what velocity must the player give the ball?

2)A baseball is hit at 32.0 m/s at an angle of 52.0° with the horizontal. Immediately an outfielder runs 5.00 m/s toward the infield and catches the ball at the same height it was hit. What was the original distance between the batter and the outfielder?

Homework Equations

D=.5at^2
t/9.8=Vy
D=Vt

The Attempt at a Solution

I tried solving the first one by using those equationswith some sin/cosine to get the height and horizontal distance but from there I got it wrong, by doing so I got 8.18m/s but is incorrect

2) I basically did the same as aboe but since there is a second part I am not sure what to do with it

If there are any other webistes with guides or examples with these types of problems please post

 

[LowlyPion]

Homework Statement

1)A basketball player tries to make a half-court jump-shot, releasing the ball at the height of the basket. Assuming the ball is launched at 53.0°, 13.5 m from the basket, what velocity must the player give the ball?

2)A baseball is hit at 32.0 m/s at an angle of 52.0° with the horizontal. Immediately an outfielder runs 5.00 m/s toward the infield and catches the ball at the same height it was hit. What was the original distance between the batter and the outfielder?

Homework Equations

D=.5at^2
t/9.8=Vy
D=Vt

The Attempt at a Solution

I tried solving the first one by using those equationswith some sin/cosine to get the height and horizontal distance but from there I got it wrong, by doing so I got 8.18m/s but is incorrect

2) I basically did the same as aboe but since there is a second part I am not sure what to do with it

If there are any other webistes with guides or examples with these types of problems please post

Show how you arrived at 8.18m/s. I got a different answer. Perhaps I can see where your method differs?

 

[Xender]

for the second problem, solve for time (which will be how long the ball is in the air for), and distance (how far the ball goes). finding time will allow you to figure out how far the outfielder moves to catch it. then just do some simple addition of distances and that is your answer (make sure the distance the ball goes u solve for the X distance)

 

[halo9909]

for the 1st one with the basketball I did
tan(53)=y/13.5m
which lead me to 18.7m
then i took 187.7m used it in D=.5at^2
which lead the "t" to be 1.912
Since i want the "horizonatal verlocity" i would be d=vt I have the d which is 13.5 now I have the time, but it is multipled by 2 to make it 3.82sec
so
13.5=V3.82
V=3.53, guess this is what I got this time.

For th 2nd one
I did 32sin53 = 25.216 yvelocity
and 32cos53 = 19.701 xvelocity

to find time I took 25.216/9.8=2.573sec then multiplied by 2 and got 5.146sec for the Horizontal
since I know the xvelocity, I did 19.701 x 5.146 and got 101.38m which is the range, the runner @ 5m/s I am still unsure of

 

[LowlyPion]

for the 1st one with the basketball I did
tan(53)=y/13.5m
which lead me to 18.7m

That's just not correct.

The Tangent of the angle of launch is not equal to the max height times the range. I would suggest that you forget that equation.

Vy = g*t
Vx = X/t

Total time = 2*Vy/g = X/Vx

 

[halo9909]

That's just not correct.

The Tangent of the angle of launch is not equal to the max height times the range. I would suggest that you forget that equation.

Vy = g*t
Vx = X/t

Total time = 2*Vy/g = Vx/X

so are my "t" correct?

 

[LowlyPion]

so are my "t" correct?

You don't need t. I didn't bother to calculate it.

Note: I made a typo in the earlier post. It should read

Total time = 2*Vy/g = X/Vx

 

[halo9909]

You don't need t. I didn't bother to calculate it.

Note: I made a typo in the earlier post. It should read

Total time = 2*Vy/g = X/Vx

Dont you need the "t" to get the Vy and Vx?

 

[LowlyPion]

Dont you need the "t" to get the Vy and Vx?

No. Look at the equation I gave you. t is eliminated.

2*Vy/g = X/Vx

 

[DDRchick]

you're also forgetting:
range= vx(time total)
if that helps any.
13.5 m would be the range...right?
To use cos, sin or tan you'd have to have a velocity (i think)...where you would get a velocity i have no idea lol.

and lowly, how would velocity be found? I mean, there's no initial velocity...only final (which would be 0 for up) ...and there's no time.
There's only range and an angle measurement...

 

[halo9909]

then how would I get the Vx /Vy
if possible could someone please show the steps, as this is getting quite confusing

 

[DDRchick]

1)
vx=cos53
vy=sin53
13.5/2 = 6.75m
6.75 going up, 6.75 going down.

t=d/v
t=6.75/cos53v

Timeup=vf-vi/a
=sin53v/9.8

6.75m/cos53v=sin53v/9.8

cross multiply and it's just algebra from there. :)

2)
sin52=y/32
cos52=x-5/32

vy/9.8=time

time(2)=total
total(x m/s) = Range :)

You just have to use a little algebra, that's all.

 

[LowlyPion]

then how would I get the Vx /Vy
if possible could someone please show the steps, as this is getting quite confusing

These are the components of initial velocity. Specifically

V_x = V*Cosθ

V_y = V*Sinθ

Putting that in the equation 2*Vy/g = X/Vx yields:

V² = g*X/(2*Sinθ*Cosθ)

Recognizing the identity 2*Sinθ*Cosθ = Sin2θ this can be simplified to:

V = (g*X/Sin2θ)^1/2