A ball is fired from a cannon at a 37 degree angle above the horizontal. There i

Question

A ball is fired from a cannon at a 37 degree angle above the horizontal. There is a tree on the edge of a cliff exactly 40 m away from the launching point. The cliff falls 20 m below the launching point and then remains level.

For this problem use g = 10 m/s2.

a) Using the information above, write equations for the x- and y-position of the ball as functions of time and the initial velocity v0.

b) If the tree is 25 m tall, at what initial velocity v0 must the ball be fired in order to just hit the top of the tree?

c) If we now cut the tree down and fire another ball with a new initial velocity of 25 m/s, but at the same angles, where will it land? (Include both an x and y position).

Please explain how you came about to a solution, I am looking more for a tutorial rather than the answers.

Explanation / Answer

(a) We know that there is no acceleration in the Horizontal or X direction therefore
X = VoCos37* t where t is the time
Y = VoSin37*t + (1/2)(-g)t2   
here initial Velocity in X direction = VoCos37
In Y direction = VoSin37
(b) If the tree is 25 m tall then at the time of hitting the ball has covere 40 m in X direction and the 5 m in Y direction
So
In X direction
40 = VoCos37*t
t = 40/V0Cos37
Now in Y direction
5 = V0Sin37*(40/VoCos37) - (1/2)(10)(40/VoCos37)2
5 = 40*tan37 - 5(40/VoCos37)2
on solving we get
Vo = 22.3367 m/s
(c) Vo = 25 m/s
time taken by the ball to reach on its level of launch
S = 0
t = 2 VoSin37/g
So horizontal distance in this much time
X = VoCos37*2VoSin37 /g = 60.07 m
but we know that after the 40 m the cliff falls down beow by 20 m
so the Y distance calculated by the ball
will be
Y = -20 = 25Sin37*t - (1/2)(10)t2
t = 4 s
X = 25Cos37*4 = 79.86 m

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