Answer to Question #81829 in Physics for Alia

Question

A hoodlum throws a stone vertically downward with an initial speed of 11.0 m/s from the roof of a building, 49.0 m above the ground. (a) How long does it take the stone to reach the ground? (b) What is the speed of the stone at impact?

Answer 1

(a) We can find the time that the stone needs to reach the ground from the kinematic equation:

y(t)=y_0+v_0 t+1/2 gt^2,

here, y(t) is the position of the stone at time t, y_0=49.0 m is the initial position of the stone, v_0=11.0 m⁄s is the initial velocity of the stone and g=-9.8 m⁄s^2 is the acceleration due to gravity.

As the stone reaches the ground, y(t)=0, and we get:

0=y_0+v_0 t+1/2 gt^2,

4.9t^2-11.0t-49.0=0.

This equation has two roots:

t_1=(11.0-√(〖11.0〗^2-4∙4.9∙(-49) ))/(2∙4.9)=-2.23,

t_2=(11.0+√(〖11.0〗^2-4∙4.9∙(-49) ))/(2∙4.9)=4.48.

Because time can’t be negative the correct answer is t=4.48 s.

(b) We can find the speed of the stone at impact from another kinematic equation:

v=v_0+gt=11.0 m/s-9.8 m/s^2 ∙4.48 s=-32.9 m/s.

The sign minus indicates that the velocity of the ball is directed downward.