Question

A stone is catapulted at time t = 0, with an initial velocity of magnitude 17.0 m/s and at an angle of 48.0° above the horizontal. (Neglect air resistance.) Find its horizontal and vertical displacements from the catapult site at the following times after launch.

(a) 1.10 s
____________? m (horizontal)
____________? m (vertical)

(b) 1.70 s
________? m (horizontal)
_________? m (vertical)

(c) 2.59 s
__________? m (horizontal)
__________? m (vertical)

Answer 1

Step-by-step explanation:

u = 17 m/s

θ = 48°

(a) t = 1.10 s

Horizontal :

vx = u cos θ = 17 Cos 48 = 11.37 m/s

Vertical:

vy = uy - gt

vy = u Sin 48 - 9.8 x 1.10

vy = 17 x 0.743 - 98 = - 85.37 m/s

(b) t = 1.70 s

Horizontal :

vx = u cos θ = 17 Cos 48 = 11.37 m/s

Vertical:

vy = uy - gt

vy = u Sin 48 - 9.8 x 1.70

vy = 17 x 0.743 - 16.66 = - 4.03 m/s

(c) t = 2.59 s

Horizontal :

vx = u cos θ = 17 Cos 48 = 11.37 m/s

Vertical:

vy = uy - gt

vy = u Sin 48 - 9.8 x 2.59

vy = 17 x 0.743 - 25.38 = - 12.75 m/s

Step-by-step explanation: