You are standing at the second-floor window of your dorm room, trying to hit a t

Question

You are standing at the second-floor window of your dorm room, trying to hit a target outside the building at ground level by tossing a ball from the window. You may have noticed before that it is surprisingly difficult to accurately hit your target in such a situation! As you will see in this problem, one reason for this difficulty is that the point where the ball (or whatever you're tossing) hits the ground is quite sensitive to the initial launch angle. Suppose you are tossing the ball from a height of 2.3 in above ground with an initial speed of 20 m/s at a downward angle of 18 degree from the horizontal. At what horizontal distance from the window will the ball hit the floor? Now suppose you launch the ball at a downward angle of 8.0 degree instead a mere 10 degrees difference, which is rather hard to control in practice. At what horizontal distance from the window will the ball now hit the floor?

Explanation / Answer

Use the equation of motion : s = ut + 0.5*at^2

For vertical motion : 2.3 = 20*sin(18 deg)*t + 0.5*9.8*t^2

So, t = 0.3 s

So, horizontal distance traveled

= 20*cos(18 deg)*0.3 = 5.71 m <----- answer

Now for the angle = 8 deg....

For vertical motion :

2.3 = 20*sin(8 deg)*t + 0.5*9.8*t^2

So, t = 0.458 s

So, horizontal distance traveled

= 20*cos(8 deg)*0.458

= 9.1 m <------ answer

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