Physics: Acat 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.

Acat runs and jumps from one roof top to another which

blondie02

11.03.2022

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

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BELLE6284

05.12.2021

14 m

Explanation:

Applying,

P.E = WH Equation 1

Where P.E = potential energy at the highest vertical height, W = weight of the toy rocket, H = Vertical height.

Note: At the highest vertical height, the kinetic energy is 0 J

Therefore,

Mechanical energy(M.E) = Potential Energy(P.E)

make H the subject of the equation

H = P.E/W Equation 2

From the question,

Given: W = 10 N, P.E = 140 J

Substitute these values in equation 2

H = 140/10

H = 14 m

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