Finding initial velocity of basketball

[jk2455]

A basketball player makes a shot from a height of 2.50 m at an angle of 30 degrees above
the horizontal. The horizontal distance between the player and the hoop is 6.00 m. The
hoop is 3.00m above the ground. Find the magnitude of the initial velocity of the ball
which will cause it to go into the basket is.

answer is 8.91m/s

Homework Statement

given:
x direction--> delta x=6m Vi=Vcos30 a=0
y direction--> delta y=0.5m Vi=Vsin30 a= -9.8

Homework Equations

v^2=Vi^2+2a*x ?( i think|)

The Attempt at a Solution

i am a bit confused a little help please help

 

[tiny-tim]

welcome to pf!

hi jk2455! welcome to pf!

(have a delta: ∆ and a degree: ° and try using the X² and X₂ icons just above the Reply box )

x direction--> delta x=6m Vi=Vcos30 a=0
y direction--> delta y=0.5m Vi=Vsin30 a= -9.8
…
v^2=Vi^2+2a*x ?( i think|)

fine so far

but you only have v_i s and a, not v_f

(and you don't want to find v_f and then eliminate it … that won't work … you want to find t and then eliminate it ),

so instead of v_f² = v_i² + 2as, you need one of the other https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations

 

[jk2455]

ok thanks ill try that
and thanks for the note on top it will make things a bit easier

(have a delta: ∆ and a degree: ° and try using the X² and X₂ icons just above the Reply box )

 

[jk2455]

thanks i got it

 

[rwisz]

I would approach this problem by breaking it down into parts and using the appropriate equations to solve for the initial velocity.

First, I would find the time it takes for the ball to reach the hoop using the equation y = y0 + v0t + 1/2at^2, where y0 is the initial height (2.50 m), v0 is the initial velocity in the y direction (unknown), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time. Solving for t, we get t = 0.319 seconds.

Next, I would use the equation x = x0 + v0t + 1/2at^2 to find the initial velocity in the x direction. Since the ball travels a horizontal distance of 6.00 m, we can set x0 = 0 and x = 6.00 m. Solving for v0, we get v0 = 18.78 m/s.

Finally, we can use the Pythagorean theorem to find the magnitude of the initial velocity. The Pythagorean theorem states that c^2 = a^2 + b^2, where c is the hypotenuse and a and b are the side lengths of a right triangle. In this case, c represents the magnitude of the initial velocity, a represents the initial velocity in the x direction (18.78 m/s), and b represents the initial velocity in the y direction (unknown). Solving for b, we get b = 8.91 m/s.

Therefore, the magnitude of the initial velocity of the ball is approximately 8.91 m/s.