How fast did you throw the ball in order to get it into the basket?

[Kotune]

Homework Statement

You are shooting for a goal in a game of basketball. You throw the ball at an angle of 55° to the horizontal. The basket is 3.0m in front of you and 1.5m above the point where the ball is released. How fast did you throw the ball in order to get it into the basket?

Homework Equations

Answer is 7ms^-1.

The Attempt at a Solution

u(H) = 3cos55° = 1.72
u(V) = 1.5sin55° = 1.23

I don't know what to do after this, please help.

 

[altamashghazi]

u should use the equation:


y=xtanθ-(gx^2)/2(ucosθ)^2

this is the locus of curve of projectile.
θ is the angle of projection from horizontal.

 

[Kotune]

u should use the equation:


y=xtanθ-(gx^2)/2(ucosθ)^2

this is the locus of curve of projectile.
θ is the angle of projection from horizontal.

Hi sorry just a few quick questions. Would x be 3? And what would be the value of u?

 

[altamashghazi]

Hi sorry just a few quick questions. Would x be 3? And what would be the value of u?

yes x is 3 and u have to calculate u. this is ur question nah.

 

[Kotune]

u should use the equation:


y=xtanθ-(gx^2)/2(ucosθ)^2

this is the locus of curve of projectile.
θ is the angle of projection from horizontal.

u = 1.5cos 55
= 0.86

y = 3tan(55) - (9.8x3^2) / 2 (0.86cos55)^2
= -172...
Thats not correct.

 

[altamashghazi]

u = 1.5cos 55
= 0.86

y = 3tan(55) - (9.8x3^2) / 2 (0.86cos55)^2
= -172...
Thats not correct.

u is initial velocity with which u throw the ball. y=1.5m. .put it in the equation and calculate 'u'

 

[Kotune]

u is initial velocity with which u throw the ball. y=1.5m. .put it in the equation and calculate 'u'

Thanks. How would you arrange it for u?
√1.5 - (3tan(55)) - (9.8x3^2) x (2(cos55)) = u?

 

[altamashghazi]

Thanks. How would you arrange it for u?
√1.5 - (3tan(55)) - (9.8x3^2) x (2(cos55)) = u?

solve this it is somewhat lengthy but u will get right answer.
[gx^2/(xtanθ-1.5)]^1/2=ucosθ

 

[Ericv_91]

Okay, although this problem involves some rather boring and tedious math, the answer is achieved by simply using the kinematic equations as all this is is a projectile motion problem.
Having quick equations like altamashghazi suggested is nice, but knowing how to get them is always the best.

Anyways, the equation you're going to want to use is d = v*t+(1/2)a*t².

You'll have to split it into components of x and y using 1.5 as d(y) and 3.0 as d(x). Furthermore you'll have to split the velocity which is one of your two unknowns (the other is t) into x and y by multiplying it by cos55 and sin 55. Once you have those filled into the equation as well as the acceleration (hint, there is no acceleration in the x direction, only in the y) then you will have 2 equations and 2 unknowns. Try to solve that yourself.