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