what is the high point of your trajectory?
Assuming you meant the cat’s jump was 60 degrees from the horizontal (need to clarify), first construct a right triangle to get the vertical component of the cat.
From this triangle we can conclude that the initial vertical speed of the cat (call it u) is sin 60 times 9.58.
(This comes from simple trigonometry).
So we know that at the highest point its speed is 0 (just before it starts accelerating again).
(in this case, 8.3 m/s)
Application of the equations of motion.
v^2 = u^2 + 2as
0 = 8.3^2 + 2 × -9.8 ×
– 68.89 = -19.6x
s = 3.5 meters
Therefore, the highest point is 3.5 meters above the ground. If not, feel free to email me.
Blair
9.58 sin(60) = 8.3 m/s
8.3^2 = 2gh
8.3^2/2g = h
68.83/19.6 = 3.511m