Physics Forces Problems

用戶20304

2013-10-10 2:40 pm

1)An airplane is flying with a velocity of 231 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.54 km, a flare is released from the plane. The flare hits the target on the ground. What is angle θ?
graph:https://scontent-b-ord.xx.fbcdn.net/hphotos-prn2/1379782_10202063904908537_1730631013_n.jpg

2) soccer ball is kicked from the ground. When it reaches a height of 5.26m, the velocity is v=[5.15i+7.27 j]m/s.
a)Determine the maximum height that the ball reaches.

b)Determine the total horizontal distance travelled by the ball.

c)What is the speed of the soccer ball just before it hits the ground?

3)A ball rolls off of a 1.50m high table. The ball hits the floor 1.50m away from the foot of the table.
a)How long did it take the ball to hit the ground?

b)Determine was the ball's initial speed when it first left the table?

回答 (1)

天同

2013-10-10 7:58 pm

✔ 最佳答案

1. Let d be the horizontal distance travelled by the flare, and t be the time of flight.
Hence, d = [231.cos(30)].t
t = d/200.1

Consider the vertical motion of the flare, use s = ut + (1/2)at^2
with s = 2540 m, u = 0 m/s, a =g(= 10 m/s^2), t = d/200.1
hence, 2540 = (1/2).(10)(d/200.1)^2
d = 4510 m
Thus, tan(θ) = 2540/4510
θ = 29.39 degrees.

2.(a) Consider the vertical motion, use v^2 = u^2 + 2as
with v = 0 m/s, u = 7.27 m/s, a =-g(= -10 m/s^2), s =?
hence, 0 = 7.27^2 + 2.(-10)s
s = 2.643 m
Thus, the height reached by the ball = (5.26 + 2.643) m = 7.903 m

(b) Use equation: s = vt - (1/2)at^2 to find the time for the ball to reach the max height
7.903 = -(1/2).(-10)t^2
t = 1.257 s
Hence, total flight time = 2 x 1.257 s = 2.514 s
Horizontal distance travelled = 5.15 x 2.514 m = 12.95 m

(c) Consider the ball's vertical falling motion, use equation v = u + at
with u = 0 m/s, a = -g(=-10 m/s^2), t = 1.257 s, v =?
hence, v = (-10).(1.257)
v = -12.57 m/s
Thus, speed of ball = square-root[(-12.57)^2 + 5.15^2] m/s = 13.57 m/s

3. (a) Use equation: s = ut + (1/2)at^2
with s = -1.5 m, u = 0 m/s, a = -g(=-10 m/s^2), t =?
-1.5 = (1/2).(-10)t^2
t = 0.5477 s

(b) Initial speed of the ball = 1.5/0.5477 m/s = 2.739 m/s

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