I. u = 6 m/s. Question 7, chap 3, sect 2 part 2 of 2 10 points 2. u 9V3 m/s. 3.
ID: 1580050 • Letter: I
Question
I. u = 6 m/s. Question 7, chap 3, sect 2 part 2 of 2 10 points 2. u 9V3 m/s. 3. u = 9V5 m/s. Note: Give the angle in degrees, use coun- terclockwise as the positive angular direction, between the limits of -180° and +180° from the positive r axis. 4. u = m/s. " u = 9V2 m/s. V5 9 9v/2 What is the direction of this resultant vec- tor F? Answer in units of °. 6. u = m/s. Question 8, chap 2, sect 5. part 1 of 1 10 points Initially (at time t = 0) a particle is moving 8, U-3V3 m/s. vertically at 4 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s 9, v 9 m/s. The acceleration of gravity is 9.8 m/s At what tine will the particle be traveling V2 9 10.- m/s. at 35 with respect to the horizontal? Answer in units of s. Question 10, chap 2, sect 8. part 1 of 1 10 points Question 9, chap 2, sect 8. part 1 of 1 10 points Denote the initial speed of a cannon ball fired from a battleship as tp. When the initial A target lies flat on the ground 9 m from projectile angle is 45° with respect to the the side of a building that is 10 m tall, as horizontal, it gives a maximum range R. shown below. resistance is negligible. roof of the building in the direction of the The acceleration of gravity is 10 m/s. Air A student rolls a 9 kg ball off the horizontal The maximum height hmar of the cannon- ball is given by V3 The horizontal speed v with which the ball mmust leave the roof if it is to strike the target 4. hmar is most nearlyExplanation / Answer
8. vx = 0 + (2.1)t = 2.1 t
and vy = 4 - (9.8)t
and tan35 = vy/vx = (4 - 9.8 t) / 2.1 t
4 - 9.8t = 1.47 t
t = 0.355 sec ....Ans
9. in vertical,
yf - yi = v0y t + ay t^2 /2
0 - 10 = 0 - 10 t^2 / 2
t = 1.414 sec
in horizontal,
x = (v0x) t { as a_x = 0 }
9 = sqrt(2) v
v = 9 / sqrt(2) Or 9 sqrt(2) /2
Ans(7)
10. Hmax = (v0 sin(theta))^2 / 2 g
= (v0 sin45)^2 / 2 g
= v0^2 / 4 g
Ans(7)