A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use
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Answer:
4.3 m/sec
Explanation:
Here height of cliff = y = 37.6 m
Gravitational acceleration = g = 9.8 m/sec2
vi = 0 m/s
Let's find the time which the diver will take if jumps from there!
Using formula
y = vit+1/2gt2
==> 37.6= 0 + 0.5 ×9.8×
==>=
==> t = 2.8 sec
In this time the diver has to cover a horizontal distance of 12.12 m
If x = 12.12 m is the horizontal distance to be covered then using
x= Vx × t
==> Vx = x/t
==> Vx= 12.12/2.8 = 4.3 m/s
a) Time of flight: 22.6 s
To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.
The vertical position at time t is given by
where
h = 2.5 km = 2500 m is the initial height
is the initial vertical velocity of the cargo
g = 9.8 m/s^2 is the acceleration of gravity
The cargo reaches the ground when
So substituting it into the equation and solving for t, we find the time of flight of the cargo:
b) 7.5 km
The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:
So the horizontal distance travelled is
And if we substitute the time of flight,
t = 22.6 s
We find the range of the cargo: