Question

Check Point: Two tall buildings face each other and are at a distance of 100 m from each other.
With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one
building so that it enters a window 10.9 m above the ground in the second building? Ans: 60 m​

Answer 1

Answer:

33.33m/s (question is wrong)

Explanation:

Difference in heights between the two windows = 55-10.9 m = 44.1 m

For horizontal motion, initial velocity, u = 0m/s

Vertical distance = $\frac{1}{2}gt^2$

=$44.1 = \frac{1}{2} X 9.8Xt^2\\\\\frac{44.1}{4.9} = t^2\\\\ t= 3 s$

horizontal distance $=xt$ (x=horizontal velocity)

$100 = 3(x)\\\\x = 100/3 = 33.33m/s$

The answer is given wrong since, if the horizontal distance had been 180 m, it wud have been the ans of 60m/s.

This method is the only method with which this sum can be solved.