Question

A basketball ball is being shot at an angle of 25 degrees with an initial velocity of 25m/s to a basketball ring (a) What is the time flight at the maximum height? (b) What is the Maximum Height? (c) What is the total time of flight? And (d) What is the range of the ball from the shooter to the ring given that it is a basket?

Answer 1

here I give step by step answer

Answer 2

Given info : A basketball ball is being shot at an angle of 25 degrees with an initial velocity of 25m/s to a basketball ring.

To find :

(a) the time of flight at the maximum height is ..

(b) the maximum height is ..

(c) the total time of flight is ..

(d) the range of the ball from the shooter to the ring given that it is a basket

solution : (a) time of flight at the maximum height , t₁ = usinФ/g

here u = initial velocity of ball = 25 m/s

Ф = angle of projection = 25°

g = acceleration due to gravity = 10 m/s²

now, t₁ = $\frac{25sin25^{\circ}}{10}$ = 1.056 sec

so the time of flight at the maximum height is 1.056 sec.

(b) the maximum height , H = $\frac{u^2sin^2\Phi}{2g}$ = $\frac{(25)^2sin^225^{\circ}}{2\times10}$ = 5.6 m

therefore the maximum height of the ball is 5.6 m.

(c) total time of flight = 2 × time of flight at the maximum height

= 2 × 1.056 = 2.112 sec

therefore the total time of flight of the projectile is 2.112 sec.

(d) the range of the ball, R = $\frac{u^2sin2\Phi}{g}$ = $\frac{25^2sin50^{\circ}}{10}$ = 47.87 m

therefore the range of the ball from the shooter to the ring given that it is a basket , is 47.87m