3. A projectile is fired from a vertical cliff edge (height = 100 m) with a horizontal velocity of 20
1 answer:
Refer to the diagram shown below.
u = 20 m/s, the horiontal launch velocity
v = 0, initial vertical velocity
Assume that wind resistance may be ignored.
The time, t, of the flight is given by
(1/2)*(9.8 m/s²)*t² = (100 m)
t² = 100/4.9 = 20.4082
t = 4.5175 s
Th horizontal distance traveled is
d = (20 m/s)*(4.5175 s) = 90.35 m
Answer: B
The projectile hits the ground at approximately 90 m from the base of the cliff.
u = 20 m/s, the horiontal launch velocity
v = 0, initial vertical velocity
Assume that wind resistance may be ignored.
The time, t, of the flight is given by
(1/2)*(9.8 m/s²)*t² = (100 m)
t² = 100/4.9 = 20.4082
t = 4.5175 s
Th horizontal distance traveled is
d = (20 m/s)*(4.5175 s) = 90.35 m
Answer: B
The projectile hits the ground at approximately 90 m from the base of the cliff.
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Answer:
33 seconds.
Explanation:
The equation for speed with constant acceleration at time t its:
where is the initial speed, and a its the acceleration.
Starting at rest, the initial speed will be zero, so
the final speed is
and the acceleration is
.
Taking all this together, we got
So, for the first half of the problem we got a time of 10 seconds.
<h3>Second half of the problem</h3>Now, the initial speed will be
,
the acceleration
,
with a minus sign cause its slowing down, the final speed will be
Taking all together:
So, for the first half of the problem we got a time of 23 seconds.
<h3>Total time</h3>