Question

2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal

Answer 1

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels x after time t horizontally is

x = (7.2 m/s) t

The object's height y at time t is

y = 9.4 m - 1/2 gt²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

v = -gt

(a) The object hits the ground when y = 0:

0 = 9.4 m - 1/2 gt²

t² = 2 * (9.4 m) / (9.80 m/s²)

t ≈ 1.92 s

at which time the object's vertical velocity is

v = -g (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

x = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m