A projectile is fired straight upward with an initial veloc- ity of 100 m=s from the top of a building 20 m high and falls to th
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Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m
Answer: It takes 2.85 seconds.
Explanation: according to the question, the kinematics equation for vertical motion is
y₀ is the initial postion and equals 0 because it is fired at ground level;
v₀ is the initial speed and eqauls 14m/s;
g is gravity and it is 9.8m/s²;
y(t) is the final position and equals 0 because it is when the pumpkin hits the ground;
Rewriting the equation, we have:
0 + 14t - = 0
14t - 4.9t² = 0
t(14 - 4.9t) = 0
For this equation to be zero,
t = 0 or
14 - 4.9t = 0
- 4.9t = - 14
t =
t = 2.86
It takes 2.86 seconds for the pumpkin to hit the ground.