Question

A cat runs and jumps from one roof top to another which is 7.22 meters away and 5.97 meters below. Calculate the minimum horizontal speed with which the cat must jump off the first roof in order to make it to the other roof.

Answer 1

Height of one roof is 5.97 m below the other roof

now its displacement in vertical direction is given as

$y = v_i t + \frac{1}{2} at^2$

$5.97 = 0 + \frac{1}{2}\times 9.8 (t^2)$

solving above for time

$t = 1.10 s$

now in the same time cat has to cover the distance between two buildings

so here we can say

$d = v_x \times t$

$7.22 = v_x \times 1.10$

solving above for speed

$v_x = 6.54 m/s$

so speed of cat is 6.54 m/s