Question

A guy playing basket-ball needs to make the penalty shot for his team to win. Given that the ball is thrown from a height of 2 meters at an angle of 60 degrees, that the post is situated at a distance of 7 m and standing at 3.5 m:

a) What does the initial velocity have to be?

b) What is the balls apogee?

c) What is the flight time to the basket?

I'm not sure what formulas to choose, etc. The method is not clear to me.

Answer 1

Since the equation of motion is

y = y₀ + xtan

Answer 2

Since the equation of motion is

y = y0 + xtanθ - gx^2/(v cosθ)^2

we have

y = 2 + √3 x - 19.6x^2/v^2

at x=7 we have
3.5 = 2 + 7√3 - 960.4/v^2
v = 9.51 m/s

So, now our equation reads

y = 2 + √3 x - 0.217x^2

the ball reaches apogee at x = √3/.434 = 4.0 m

y = 5.456m


Since the horizontal velocity = v cosθ = 4.755m/s, it takes 7/4.755 = 1.467 seconds to fly

Answer 3

so it's always a tan when it's 2 dimensions? Thank you very much

Answer 4

typo. That's

gx^2/2(v cosθ)^2

but the numbers are right. (I hope)