{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

qz2sol_0132Hf10

# qz2sol_0132Hf10 - a What is the magnitude of the ball’s...

This preview shows page 1. Sign up to view the full content.

TA: Tomoyuki Nakayama Monday, September 13, 2010 PHY 2048: Physic 1, Discussion Section 0132H Quiz 2 (Homework Set #3) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ A batter hits a pitched ball when the center of the ball is 1.20 m above the ground. The ball leaves the bat an angle of 30.0º with the ground. With that launch, the ball should have a horizontal range of 105 m. A 5.00-m high fence is located 95.0 m horizontally form the launch point.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a) What is the magnitude of the ball’s initial velocity? The equation for horizontal range yields R = (v 2 /g)sin2 θ ⇒ v = √ (Rg/sin2 θ ) = 34.5 m/s b) At the fence, what is the distance between the fence top and the ball center? The x and y components of the initial velocity are v 0x = v cos θ = 29.9 m/s, v 0y = v sin θ = 17.3 m/s The ball reaches the position of the fence at Δ x = v 0x t t = Δ x/v 0x = 3.18 s At that moment, the height of the ball is y(t = 3.18s) = y + v 0y t – (1/2)gt 2 = 6.66 m Therefore, it is 1.66 m (= 6.66 – 5) above the fence....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online