Question

A rock is hurled horizontally from the top of a 55 m building and lands 145 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive xin the direction of the throw.a) How long is the rock in the air in seconds.b) What must have been the initial horizontal componant of the velocity, in meters per second?c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Answer 1

Answer:

a) 3.35 s

b) 43.4 m/s

c) 32.85 m/s

d) 54.35 m/s

Explanation:

a) If we ignore air resistance, then gravitational acceleration g = 9.81 m/s2 is the only thing that generate vertical motion of the rock. We can use the following equation of motion to find out the time it travels 55m vertically with acceleration g = 9.81m/s2 and initial vertical speed = 0 m/s

$h = gt^2/2$

$55 = 9.81t^2/2$

$t^2= \frac{55*2}{9.81} = 11.2$

$t = \sqrt{11.2} = 3.35 s$

b) Since there's no air resistance, the rock horizontal velocity stays constant. For it to travel 145m horizontally within 3.35 s, its horizontal velocity must be

$v_h = 145 / 3.35 = 43.3 m/s$

c) The vertical component of the rock velocity just before it hits the ground is the product of acceleration g and the time t

$v_v = 3.35 * 9.81 = 32.85 m/s$

d) The magnitude of the rock velocity of the rock before it hits the ground consists of both horizontal and vertical velocity

$v = \sqrt{v_h^2 + v_v^2} = \sqrt{43.3^2 + 32.85^2} = \sqrt{1874.89 + 1079.1225} = \sqrt{2954.0125} = 54.35 m/s$