what is the highest point of its trajectory?

Assuming you mean the cats jump was 60 degrees from the horizontal (needs clarified) then firstly construct a right angled triangle to get the vertical component of the cat.

From this triangle we can conclude that the cats initial vertical velocity (lets call it u) is sin 60 times 9.58.

(This comes from simple trig).

Then we know at the highest point its velocity is 0 (just before it starts accelerating back down).

(in this case, 8.3 m/s)

Applying the equations of motion.

v^2 = u^2 + 2as

0 = 8.3^2 + 2 x -9.8 x s

– 68.89 = -19.6 x s

s = 3.5 meters

Hence the highest point is 3.5 meters above ground. If you haven’t followed feel free to e-mail me.

Blair

9.58sin(60) = 8.3 m/s

8.3^2 = 2gh

8.3^2/2g = h

68.83/19.6 = 3.511 m