A cat jumps off a table which is 1.5 m high. If the initial velocity of the cat is 10 m/s, at an angle of 37 degrees above the horizontal, how far from the edge of the table does the cat land?

11.5 m

Explanation:

First, find the time it takes to land.

Given:

Δy = - 1.5 m

v₀ = 10 sin 37° m/s

a = - 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-1.5 m = (10 sin 37° m/s) t + ½ (-9.8 m/s²) t²

-1.5 = 6.02 t - 4.9 t²

4.9 t² - 6.02 t - 1.5 = 0

Solve with quadratic formula:

t = [ - (-6.02) ± √ ((-6.02) ² - 4 (4.9) (-1.5)) ] / 2 (4.9)

t = 1.44

Now find how far it moves horizontally in that time:

Given:

t = 1.44 s

v₀ = 10 cos 37° m/s

a = 0 m/s²

Find: Δx

Δx = v₀ t + ½ at²

Δx = (10 cos 37° m/s) (1.44 s) + ½ (0 m/s²) (1.44 s) ²

Δx = 11.5 m

The cat lands 11.5 meters from the table.