Projectile Motion question (1 Viewer)

[gordon.tan2008]

A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a height double that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity of the stone.


pg 124 Fitzpatrick 3U green book q.17

 

[vds700]

A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10m/s. The stone reaches a height double that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity of the stone.


pg 124 Fitzpatrick 3U green book q.17

here u go

 

[lolokay]

http://community.boredofstudies.org...op+of+pole.+However,+instant+stone+is+thrown,

t1 = time for rock to reach pole
t2 = time from pole to max height
t3 = time from max height to pole height, which is time it hits bird, = t2
t.t = total time = t(1+2+3)
p = pole height
r = square root pole height
d = distance from initial point of rock to pole (horizontal distance)

Using equations v^2 = u^ + 2as and v = u + at you get approximate values of t(1+2) = 6.26r/9.8 t3 = 4.43r/9.8, so t1 = 1.83r/9.8, t2 = 4.43r/9.8, t.t = 10.69r/9.8.

0 = u^2 + 2*-9.8*2p
u ^2 = 39.2p
u = 6.26r
0 = 6.26r - 9.8t
t = 6.26r/9.8
this is section t(1+2)

v^2 = 0 + 2*-9.8*-p
v^2 = 19.6p
v = 4.43r
4.43r = 0 + 9.8t
t = 4.43r/9.8
this is section t3


The distance the bird has travelled is 10t.t = (106.9r)/9.8.
The horizontal distance the stone has to travel is (106.9r)/9.8 + d.
The horizontal velocity is constant, so v = d/t1: d = (1.83r*v)/9.8

horizontal velocity = horizontal distance / time
v = (106.9r/9.8 + 1.83rv/9.8)/(10.69r/9.8)
10.69rv = 106.9r + 1.83rv
8.86rv = 106.9r
v = 12.065 m/s

since the values were approximations it probably is closer to 12.1 (the given answer)

PS diagrams really help