A rock is thrown horizontally from a bridge with a velocity of 13.0 m/s. It takes the rock 3.5 s to strike the water below. What is the magnitude of the velocity of the rock just before it hits the water?
vertical velocity=g*3.5
horizonal velocity=13
velocity=sqrt(13^2+(g*3.5)^2)
To find the magnitude of the velocity of the rock just before it hits the water, we need to break down the problem into two components: the horizontal and vertical motions.
Since the rock is thrown horizontally, its initial vertical velocity is zero. Therefore, the only force acting on the rock in the vertical direction is gravity.
The time it takes for the rock to hit the water, 3.5 seconds, is the same for both the horizontal and vertical motions.
To find the magnitude of the velocity just before the rock hits the water, we can use the horizontal motion.
Given:
Initial horizontal velocity (Vx) = 13.0 m/s
Time (t) = 3.5 s
Using the formula for horizontal motion, which is:
Distance (d) = Vx * t
We can calculate the horizontal distance traveled by the rock:
d = Vx * t
d = 13.0 m/s * 3.5 s
d = 45.5 m
Now that we know the horizontal distance traveled by the rock, we can use this information to calculate the magnitude of its velocity just before hitting the water.
Since there is no horizontal force acting on the rock, its horizontal velocity remains constant throughout the motion. Therefore, the magnitude of the velocity just before it hits the water is equal to its initial horizontal velocity.
Therefore, the magnitude of the velocity of the rock just before it hits the water is 13.0 m/s.