A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil
2 answers:
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Answer:
The height will be 4 times.
Explanation:
Given that,
The speed at the bottom of the hill doubled.
We need to calculate the height
Using conservation of energy
Therefore,
Here, m and g are constant
Hence, The height will be 4 times.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
= 4.7476 m/sec
= 4.75 m/s
