May
SHOW YOUR WORK IN ALL QUESTIONS. Acceleration due to gravity: a = 9.81 m/s2
1.) (a) A ball is dropped from 4 meters above the ground. If it begins at rest, how long does it take
to hit the ground?
(b) A ball is thrown upward at 4 meters per second starting from ground level. How long does it
take for the ball to return to the ground?
(c) If a ball that is 4 meters above the ground is thrown horizontally at 4 meters per second,
how long will it take for the ball to hit the ground?
(d) In question (c) above, how far will the ball travel in the horizontal direction before it hits th
ground?
(e) Drop a ball from a height of 2 meters and, using a stopwatch, record the time it takes to
reach the ground. Repeat this two more times and record all the times in the table belov-
then find the average time. Be sure to release the ball from rest rather than throwing it
or down.
Test Number
Time (seconds)
Answer:
a) t = 0.90 s, b) t = 0.815 s, c) t = 0.90 s, d) x = 3.6 m, e) t = 0.639 s
Explanation:
all these exercises are about kinematics
a) The body is released from rest,
y = y₀ + v₀ t – ½ g t²
in this case when reaching the ground y = 0 and its initial velocity is vo = 0
0 = y₀ + 0 – ½ g t²
t² = 2 y₀ / g
t² = 2 4 /9.81
t² = 0.815
t = √0.815
t = 0.90 s
b) It is thrown upwards at v₀ = 4 m / s
y = y₀ + v₀ t – ½ g t²
in this case the initial and final height is the same
y = y₀ = 0
0 = v₀ t -1/2 g t²
t = 2 v₀ / g
t = 2 4 /9.81
t = 0.815 s
c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s
y = y₀ + v_{oy} t – ½ g t²
0 = y₀ + 0 – ½ g t²
t² = 2 i / g
t² = 2 4 / 9.81
t² = 0.815
t = 0.90 s
d) the horizontal distance traveled is
x = v₀ₓ t
x = 4 0.90
x = 3.6 m
e) We can calculate the time to fall from I = 2 m
y = y₀ + v_{oy} t – ½ g t²
0 = y₀ + 0 – ½ g t²
t² = 2 y₀i / g
t² = 2 2 /9.81
t² = 0.4077
t = 0.639 s
Therefore, when making measurements, you should find readings around this value.