Please show your work.Jud Rice said:Homework Statement
A rock is thrown off of a 32 m high building with a horizontal velocity of 5 m/s. What is the vertical velocity when it reaches 6 m from the ground? Use g = 10 m/s^2
A) 11 m/s
B) 120 m/s
C) 23 m/s
D) 520 m/s
E) 16 m/s
Homework Equations
Kinematics I think
The Attempt at a Solution
I think it is 23 m/s but not sure
No.berkeman said:Please show your work.
The vertical velocity of a projectile can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity (in this case, 0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time the rock takes to reach the ground. In this scenario, the rock will take approximately 1.6 seconds to reach the ground, and therefore, the vertical velocity will be -15.68 m/s.
The height of the building does not affect the vertical velocity of the rock, as long as the initial height (32m) and the final height (6m) are constant. The only factors that affect the vertical velocity are the acceleration due to gravity and the time it takes for the rock to reach the ground.
The initial velocity of the rock is considered to be 0 m/s because it is thrown from a stationary position (i.e. it is not already moving at the time of release). In other words, the rock starts with no initial velocity, and the only force acting on it is the force of gravity.
Yes, the vertical velocity of the rock can be positive if it is thrown upwards with enough initial velocity to overcome the acceleration due to gravity. However, in this scenario, the rock is thrown downwards, and therefore, the vertical velocity is negative.
This calculation is relatively accurate, assuming that the initial height, final height, and acceleration due to gravity are all constant. However, it is important to note that factors such as air resistance and variations in the acceleration due to gravity (due to location or altitude) may affect the accuracy of the calculation.