Question

. A new game has been invented where the player has to combine a dive from a board with shooting of a basketball through a hoop (see figure below). As a student of PHY 101 you decide to take part in this game. During your practice sessions, you realize that while standing on the diving board 4.50 m above the water, you are able to score into the hoop which is a horizontal distance of 6.50 m away from the board and 1.00 m above ground level. (a) With what speed did you launch the basketball? (b) Now you step off the diving board and launch the ball in order to make a basket into a hoop that is a horizontal of 3.75 m, at what time after stepping off the board should you launch the ball? (c) What are the horizontal and vertical components of the ball’s velocity at the instant of launch?

Answer 1

Step-by-step explanation:

Not being able to see the picture attached and supposing the ground level is considered to be the water level so to be able to find out the velocity you need to score into the hoop you need to use the combined force ( resulting force) to throw the ball at 6.50 m distance, 1 m below to where you are standing

Answer 2

a. 7.69 m/s b. 0.357 s c. initial vertical velocity component at launch u = 3.50 m/s. initial horizontal velocity component at launch u₁ = 7.69 m/s

Step-by-step explanation:

a. Let H be the distance of the board above the water = 4.50 m and h be the distance of the hoop above the ground = 1.00 m.

Using s = ut - 1/2gt² with u = 0 (the initial vertical component of the velocity), s = vertical distance dropped by ball = h - H = 1.00 - 4.50 = -3.50

s = 0 - 1/2gt² = -1/2gt²

t = √(-2s/g) = √(-2 × -3.50/9.8) = √(7.0/9.8) = √0.7143 = 0.845 s.

This is the time it takes the ball to enter the hoop.

The speed with which it covers the 6.50 m horizontal distance is in this time is

v = d/t = 6.50/0.845 = 7.69 m/s

b. The time taken for the ball to enter the horizontal hoop at a distance of 3.75 m is t₁ = d/v = 3.75/7.69 = 0.4875 s ≅ 0.488 s

The time t₂ after stepping off the board is t₂ = t - t₁ = 0.845 - 0.488 = 0.357 s

c. For the initial vertical velocity component at launch u, we use v = u - gt. Since the final vertical velocity v = 0, we have 0 = u -gt ⇒ u = gt where t = time it took for first throw on diving board = 0.357 s

u = gt = 9.8 × 0.357 = 3.4986 m/s ≅ 3.50 m/s

The initial horizontal velocity component at launch u₁ = v = 7.69 m/s