Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

A basketball player scores a three point shot form the three point line, which is 7.24 meters from the hoop. The ball was thrown from a height of 2.00 meters above the court at an angle of 45 degrees. The hoop is 3.00 meters above the court and air resistance can be ignored.

a) What speed was the ball thrown at?

b) How long did the ball take to reach the hoop?

c) What were the ball's velocity components when it reached the hoop?

Relevance

Let u = initial velocity f the ball,

then horizontal component, ux = ucos45 =  u/√2

and vertical component, uy = usin45 = u/√2

Horizontal travel of ball

d = (ux)t

7.24 =  ut/√2  .....(1)

Vertical travel of the ball

s = (uy) t + ½gt²

(3.00 - 2.00) = ut/√2 + ½(-9.81)t²

1.00 =  ut/√2 - 4.905t²

plug in from (1)

1 = 7.24 - 4.905t²

4.905t² = 6.24

t = 1.13 s

From (1)

u = (7.24√2)/t

u = (7.24√2)/(1.13)

u = 9.06 m/s

vertical component of velocity when ball reaches the hoop = vy

vy = uy+at

vy = (u/√2) + (-9.8)(1.13)

vy = (9.06/√2 ) - (9.8)(1.13)

vy = -4.67 m/s  ( 4.67 m/s downwards)

horizontal component of velocity when ball reaches the hoop = vx

vx = ux = u/√2 = 9.06/√2 = 6.41 m/s

(a) Speed at which ball was thrown : 9.06 m/s

(b) Time to reach the hoop : 1.13 s

(c)  Ball's velocity component at the hoop:

horizontal : 6.41 m/s

vertical : 4.67 downwards

• since the angle is 45°, then both sine e cosine are worth 0.707

vertical motion equation

(3-2) = Vo*0.707*t-4.903*t^2  (1)

horizontal motion equation

7.24 = Vo*0.707*t   (2)

t = 7.24 / (Vo*0.707)

back to equation (1) :

1 = Vo*0.707*7.24 / (Vo*0.707) - 4.903*7.24^2/(0.5*Vo^2)

257/(0.5V^2) = 6.24

257 = 3.12Vo^2

Vo = √ 257/3.12 = 9.08 m/sec

t = 7.24/(9.08*0.707) = 1.128 sec

tup = Vo*0.707/g = 9.08*0.707/9.806 = 0.655 sec

t down = t-tup= 1.128 - 0.655 = 0.473 sec

Vdown = g*t down = 0.473*-9,806 = -4.64 m/sec

Vx = Vo*0.707 = 9.08*0.707 = 6.42 m/sec