Answer:
t = 0.96 s is the time it takes for the angle to reduce
Explanation:
In the launch of projectiles, the velocity is broken down into its x and y components, the velocity in the x-axis is constant, as there is no acceleration, instead the velocity in the axis and is reduced by the effect of the acceleration of gravity.
We can find Vox for the initial conditions
Voy = Vo sin θ
Voy = 29 sin 36
VoY = 17 m/s
Vox = Vo cos θ
Vox = 29 cos 36
Vox= 23.5 m/s
Vx = Vox
The velocity on the x axis is constant
By trigonometry, we find the firing angle
tan θ = Voy/ Vx
Vy = Vx tan θ
Vy = 23.5 tan 18
Vy = 7.64 m/s
Now that we have the vertical speed we can find the time
Vy = I'm going - g t
t = (Vy -Voy) / g
t = (17 - 7.64) /9.8
t = 0.96 s
Be the time it takes for the angle to reduce
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Answer:
To achieve the velocity of 40 m/sec height will become 4 times
Explanation:
We have given initially truck is at rest and attains a speed of 20 m/sec
Let the mass of the truck is m
At the top of the hill potential energy is mgh and kinetic energy is 
So total energy at the top of the hill 
At the bottom of the hill kinetic energy is equal to and potential energy will be 0
So total energy at the bottom of the hill is equal to 
Form energy conservation 
, for v = 20 m/sec
Squaring both side
h = 20.408 m
Now if velocity is 0 m/sec
h = 81.63 m
So we can see that to achieve the velocity of 40 m/sec height will become 4 times
