Answer to Question #80694 in Physics for Rhiannon

Question

1. Consider a freely falling object.
a. What is the acceleration (in m/s2) after 5 seconds of fall?
i. 9.8 m/s^2
b. What is the acceleration (in m/s2) after 10 seconds of fall?
i. 9.8m/s^2
c. What is the velocity (in m/s) after 5 seconds of fall?
i. 49m/s
d. What is the velocity (in m/s) of 10 seconds of fall?

Answer 1

(a)-(b) The acceleration of a freely falling object is constant and equal g=-9.8 m⁄s^2 .
(c) We can find the velocity after 5 seconds of fall from the kinematic equation:
v(t=5 s)=v_0+at=0 m/s+(-9.8 m/s^2 )∙5 s=-49 m/s.
The sign minus indicates that the velocity of the object is directed downward.
(d) Similarly, we can find the velocity after 10 seconds of fall:
v(t=10 s)=v_0+at=0 m/s+(-9.8 m/s^2 )∙10 s=-98 m/s.
The sign minus indicates that the velocity of the object is directed downward.
Answer:
(a)-(b) g=9.8 m⁄s^2 , downward.
(c) v(t=5 s)=49 m⁄s, downward.
(d) v(t=10 s)=98 m⁄s, downward.