Finding Initial Velocity with no time or other Velocities

[Hypnos_16]

Homework Statement

An Olympic basketball player shoots towards a basket that is 5.52 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.52 m above the floor at an angle of 32.0o above the horizontal. What initial speed should she give the ball so that it reaches the basket and hopefully scores?

all i can get out of this is
dx = 5.52m
dy = 3.05 though she's only shooting 1.53m to the basket
Theta = 32 degs
ay = -9.81

Homework Equations

I don't know how to get the time out of this problem, nor the velocity
any time i try I'm told it would be easier to merge two equations, though i don't know how.
Please someone help.

The Attempt at a Solution

 

[tiny-tim]

Welcome to PF!

Hi Hypnos_16! Welcome to PF!

(have a theta: θ )

Call the speed v, and use the x components to find t.

Then use that value of t in the y components to find how high the ball is.

Show us what you get. ​

 

[Hypnos_16]

so I've tried writing out the formula. i got this
x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1

but i don't know what to do from there, i don't know even if that's right, and if it is where do i plug that into?

 

[tiny-tim]

so I've tried writing out the formula. i got this
x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1

(try using the X₂ icon just above the Reply box )

good so far, now your v₁ in that equation has to be the initial component of velocity in the x direction, which is v times … ?

 

[Hypnos_16]

times v by cos of 32 degs! then you get t with a value v that you can use to find v1! okay i think i get it now

t = 5.52/v1cos32
t = 6.51v1

x = x1 + v1t
5.51 = 0 + v1 x 6.51v1
5.51 = 6.51v1^2
v1 = 0.92m/s

i got this, but it doesn't seem right.
i then got a time of 6 seconds, which seems to high.

 

[tiny-tim]

times v by cos of 32 degs!

That's it!

If you have any difficulty with the rest of the problem, just come back and tell us.

 

[Hypnos_16]

so i got the v in the x direction as 0.92m/s the time in the air to be 6 seconds but the v in the y direction seems outrageously high. I think i goofed up somewhere.

x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1
t = 5.52/v1cos32
t = 6.51v1

x = x1 + v1t
5.51 = 0 + v1 x 6.51v1
5.51 = 6.51v1^2
v1 = 0.92m/s

t = 6.51(0.92)
t = 6s

dy = y1 + v1t -gt^2
3.05 = 1.53 + sin32v1(6) - 4.9(36)
1.52 = sin32v1(6) - 176.4
177.92 = sin32v1(6)
177.92 = 3.18v1
55.9 = v1

 

[tiny-tim]

x = x1 + v1t
5.52 = 0 + v1t
5.52 = v1t
t = 5.52/v1
t = 5.52/v1cos32
t = 6.51v1

No, the last two lines should be
t = 5.52/vcos32
t = 6.51/v, shouldn't it?

(where v is the total initial speed)

Now use that value of t (ie use 6.51/v ) in an equation for the y direction.

 

[Hypnos_16]

so after i fill that in for t do i have to work out the result as a quadratic equation?
dy = vy1 + v1t - gt^2
3.05 = 1.53 + sin32(6.51v) - 4.9(6.51v)^2
1.52 = 3.45v - 207.7v^2
0 = 1.52 + 3.45v - 207.7v^2
that's the answer i got.
would finding this result in finding the initial overall speed or are there more steps?

 

[tiny-tim]

(try using the X² icon just above the Reply box )

dy = vy1 + v1t - gt^2
3.05 = 1.53 + sin32(6.51v) - 4.9(6.51v)^2
1.52 = 3.45v - 207.7v^2
0 = 1.52 + 3.45v - 207.7v^2

No, it's 6.51/v, not 6.51 times v.

(and i think you got your 1.53 and 1.52 the wrong way round)

would finding this result in finding the initial overall speed or are there more steps?

we defined v as the initial overall speed, didn't we?

 

[Hypnos_16]

so how would you multiply sin32(6.51/v) and 4.9(6.51/v)^2?

 

[tiny-tim]

so how would you multiply sin32(6.51/v) and 4.9(6.51/v)^2?

uhh?

it's just (sin32*6.51)/v and (4.9*(6.51)²)/v².

get some sleep! :zzz:​

 

[Hypnos_16]

will do, after i get this one done
so after all that work i got 11.7m/s
does that seem like it would be the right answer?

 

[tiny-tim]

will do, after i get this one done
so after all that work i got 11.7m/s
does that seem like it would be the right answer?

erm … I don't mind checking your calculations, but I'm not going to do the work myself!

 

[Hypnos_16]

No that's fine you don't have to do it for me, i just want to make sure I've done it correctly, make sure the math is right.

 

[tiny-tim]

No that's fine you don't have to do it for me, i just want to make sure I've done it correctly, make sure the math is right.

Well, the general technique is right, but you keep making mistakes like v instead of 1/v, so it's difficult to say without seeing the whole thing.

I'm off to bed now, anyway …

goodnight! :zzz:​