Homework #5 - Projectiles, Net Force, Newton's 2nd Law-solutions

# Homework #5 - Projectiles, Net Force, Newton's 2nd Law-solutions

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sanchez (es25396) – Homework #5 - Projectiles, Net Force, Newton’s 2nd Law – guevara – (130101) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A basketball player 2 . 02 m tall wants to make a basket from a distance of 10 . 1 m. The hoop is at a height of 3 . 05 m. The acceleration of gravity is 9 . 8 m / s 2 . Neglect air friction. 2 . 02 m 3 . 05 m 10 . 1 m v 24 . 7 ◦ If he shoots the ball (from a height of 2 . 02 m) at a 24 . 7 ◦ angle, at what initial speed must he throw the basketball so that it goes through the hoop without striking the back- board? Correct answer: 12 . 9422 m / s. Explanation: Given : h i = 2 . 02 m , h f = 3 . 05 m , and Δ x = 10 . 1 m . The components of the initial velocity are v ix = v x = v i cos θ and v iy = v i sin θ . Since Δ x = v x Δ t , the time for the ball to move Δ x horizontally is Δ t = Δ x v x = Δ x v i cos θ . At this time, the vertical displacement of the ball must be Δ y = v iy Δ t- 1 2 g (Δ t ) 2 = v i sin θ Δ x v i cos θ- g (Δ x ) 2 2 v i 2 cos 2 θ = Δ x tan θ- g (Δ x ) 2 2 v i 2 cos 2 θ Thus Δ y- Δ x tan θ =- g (Δ x ) 2 2 v i 2 cos 2 θ 2 v i 2 cos 2 θ = g (Δ x ) 2 Δ x tan θ- Δ y v i 2 = g (Δ x ) 2 2 cos 2 θ (Δ x tan θ- Δ y ) . Since Δ x tan θ- Δ y = (10 . 1 m) tan 24 . 7 ◦- (1 . 03 m) = (4 . 64548 m)- (1 . 03 m) = 3 . 61548 m , we obtain v i = radicalBigg g (Δ x ) 2 2 cos 2 θ (Δ x tan θ- Δ y ) = radicalBigg (9 . 8 m / s 2 ) (10 . 1 m) 2 2 (cos 24 . 7 ◦ ) 2 (3 . 61548 m) = 12 . 9422 m / s 002 10.0 points A ball is thrown straight upward and returns to the thrower’s hand after 3 s in the air. A second ball is thrown at an angle of 31 ◦ with the horizontal....
View Full Document

## This note was uploaded on 07/30/2009 for the course PHYS 1301 taught by Professor Guevara during the Spring '09 term at Texas A&M.

### Page1 / 5

Homework #5 - Projectiles, Net Force, Newton's 2nd Law-solutions

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online