A rock is thrown horizontally at 10.0m/s off of a 150m sheer cliff. What is the range?
2 answers:
You might be interested in
Answer:
l= 3.002 cm
Explanation:
Given that
n= 70 turns
B= 1.2 T
θ= 15°
I= 1.5 A
τ = 0.0294 N⋅m
Lets take length of sides is l.
We know that
τ = n I A B sin θ
Area of square ,A= l²
Now by putting the value
τ = n I A B sin θ
0.0294 = 70 x 1.5 x l² x 1.2 x sin 15°
l² = 0.000901 m²
l² = 9.01 cm²
l= 3.002 cm
Answer: 815.51 m
Explanation:
This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:
(1)
(2)
Where:
is the final velocity of the bullet
is the initial velocity of the bullet
is the acceleration due gravity, always directed downwards
is the time
is the vertical position of the bullet at
Let's begin by finding from (1):
(3)
(4)
Now we have to substitute (4) in (2):
(5)
Isolating :
This is the displacement of the bullet after 6.9 s