F.5 physics projected motion

YTC

2014-09-09 7:48 am

A shell is fired from a point on a horizontal ground 100m from the foot of a vertical
tower of height 90m. The angle of projection is adjusted to be 60° Such that the
shell can just clear the tower. When the shell reaches its maximum height, it
explores into 3 equal fragments. Immediately after explosion, one fragment is
found to be stationary, one fragment moves upward with a speed of 10m/s.
Use g=10m/s^2

a) find the speed of projection of the shell
b) find the maximum height reached by the shell

Picture:
http://postimg.org/image/4x5joambn/
Thank you for help!

回答 (1)

天同

2014-09-09 8:37 am

✔ 最佳答案

(a) Let U be the speed of projection of the shell.
Horizontally, [U.cos(60)].t = 100
where t is the time of flight of the shell from projection point to the top of the tower.
Hence, t = 100/[U.cos(60)]

Vertically, use equation: s= ut + (1/2)at^2
with s = 90 m, u = U.sin(60), a = -g(=-10 m/s^2), t = 100/[U.cos(60)]
hence, 90 = [U.sin(60)].[100/(U.cos(60)] + (1/2).(-10).[100/(U.cos(60)]^2
i.e 90 = 100.tan(60) - 5[10000/U^2.cos^2(60)]
solve for U gives U = 49.03 m/s

(b) Use equation: v^2 = u^2 + 2a.s
with v = 0 m/s, u = 49.03.sin(60) m/s = 42.46 m/s, a = -g(=-10 m/s^2), s =?
hence, 0 = 42.46^2 + 2(-10).s
s = 90.14 m

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