Quiz1

# Quiz1 - Phys 2101 Sec6on 7 ...

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

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

Unformatted text preview: Phys 2101 Sec6on 7  Quiz 1 – Form A   Aug 31 2010  A plane, diving with constant speed at an angle of 60° with the ver8cal, releases a  projec8le at an al8tude of 700 m. The projec8le hits the ground 6 s a?er release.  (a)  What is the speed of the plane?   In the drawing above, the angle of the ini8al velocity vector with the posi8ve  horizontal x‐axis is θ=-30o, and the ini8al height is y0=700m.  We know that at  t=6s, the projec8le is on the gorund, which we will take as y=0.   The equa8on for the y‐component of the posi8on vector is   1 y = y 0 + v 0 sin θ t − gt 2 2 so at t=6s, when y=0, we know everything but the ini8al speed:    1 0 = y 0 + v 0 sin θ t − gt 2 ⇒ € 2 12 gt − y 0 0.5 × 9.8 m / s2 × (6 s) 2 − 700 m −523.6 m 2 v0 = = = = 174.53 m / s sin θ t sin(−30 o ) × 6 s −3s € (No8ce that we cannot get a nega8ve number for the ini8al speed unless we do  something wrong…)  (b) What is the projec6le’s speed when it hits the ground?   We need x and y components of the velocity at t=6s:  v x = v 0 cosθ = 174.53 m/s × cos(−30 o ) = 151.15 m/s v y = v 0 sin θ − gt = 174.53 m/s × sin(−30 o ) − 9.8m/s2 × 6 s = −146.07 m/s v = v x 2 + v y 2 = 210.19 m/s € Phys 2101 Sec6on 7  Quiz 1 – Form B   Aug 31 2010  A plane, diving with constant speed at an angle of 50° with the ver8cal, releases a  projec8le at an al8tude of 500 m. The projec8le hits the ground 5 s a?er release.  (a) What is the speed of the plane?   In the drawing above, the angle of the ini8al velocity vector with the posi8ve  horizontal x‐axis is θ=-40o, and the ini8al height is y0=500m.  We know that at  t=5s, the projec8le is on the ground, which we will take as y=0.   The equa8on for the y‐component of the posi8on vector is   1 y = y 0 + v 0 sin θ t − gt 2 2 so at t=5s, when y=0, we know everything but the ini8al speed:   € 1 0 = y 0 + v 0 sin θ t − gt 2 ⇒ 2 12 gt − y 0 0.5 × 9.8 m / s2 × (5 s) 2 − 500 m −377.5 m v0 = 2 = = = 117.6 m / s sin θ t sin(−40 o ) × 5 s −3.21s (b) How far does the projec8le travel horizontally during its ﬂight?  € Since we know the projec8le travels with constant velocity in the horizontal  direc8on, we know that distance = velocity * 8me:   d = v 0 x t = v 0 cos θ t = 117.6 m / s × cos(−40 o ) × 5 s = 450 m € ...
View Full Document

## This note was uploaded on 12/01/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online