A rock is thrown vertically up into the air. It reaches a maximum height 15m above point where it was released. What is its initial velocity?
a) 17 m/s
b) 26 m/s
c) 14 m/s
d) 20 m/s
I know the answer is a) 17 m/s and I got it wrong thinking it was d) 20 m/s but I can't seem to figure out how to find the answer. Am I using the correct formula? Vi = Vf - (a x t)?
Ignoring air resistance, equate potential and kinetic energies:
mgh = (1/2)mv²
Cancel m,
v=√(2gh)
=√(2*9.8*15)
= 17.1 m/s
To find the initial velocity of the rock, you need to use the correct formula and apply the appropriate concept of motion.
In this case, since the rock is thrown vertically up into the air and reaches a maximum height, we can assume that it experiences a constant acceleration due to gravity acting in the downward direction throughout its motion.
The correct formula to use in this scenario is:
Vf^2 = Vi^2 + 2aΔy
Where
Vf = final velocity (which is 0 m/s at the maximum height since the rock momentarily comes to rest)
Vi = initial velocity
a = acceleration due to gravity (approximately -9.8 m/s^2 since it acts in the opposite direction of the motion)
Δy = change in vertical displacement (15 m)
Plugging in the given values:
0 = Vi^2 + 2(-9.8)(15)
Simplifying the equation:
0 = Vi^2 - 294
Now, to solve for Vi:
Vi^2 = 294
Vi ≈ √294
Vi ≈ 17.14 m/s
Therefore, the correct answer is a) 17 m/s.
To find the initial velocity of the rock, you need to use the appropriate equation of motion. In this case, since the rock is thrown vertically and reaches a maximum height before falling back down, you can use the following equation:
Vf^2 = Vi^2 + 2as
Here, "Vf" represents the final velocity, "Vi" represents the initial velocity, "a" represents the acceleration (which is equal to the acceleration due to gravity, approximately 9.8 m/s²), and "s" represents the displacement (which in this case is the maximum height, 15m).
When the rock reaches its maximum height, its final velocity would be 0 m/s since it momentarily stops before falling back down. Plugging in the values into the equation, you have:
0^2 = Vi^2 + 2 * (-9.8) * 15
Simplifying this equation, you get:
0 = Vi^2 - 294
Rearranging the equation, you can solve for Vi:
Vi^2 = 294
Taking the square root of both sides, you find:
Vi = ±√294
Since velocity is a scalar quantity (magnitude without direction), you disregard the negative sign and obtain:
Vi ≈ 17 m/s
So the correct answer is indeed a) 17 m/s. Your initial formula, Vi = Vf - (a x t), is not applicable in this scenario since time (t) is not explicitly given. Instead, you need to use the equation based on the principles of motion and displacement.