You throw a ball horizontally off a building with a horontal velocity of 10 m/s. What is the ball's horontal velocity 5 seconds later? What is the ball's vertical velocity 5 seconds later?
horizontal velocity is constant.
vertically after t seconds, v = -9.8t
To find the ball's horizontal velocity after 5 seconds, we can assume that there is no horizontal acceleration, and the initial velocity remains constant.
Therefore, the ball's horizontal velocity after 5 seconds is still 10 m/s.
To find the ball's vertical velocity after 5 seconds, we can use the formula for vertical motion under constant acceleration:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration (due to gravity, -9.8 m/s^2)
t = time (5 seconds)
Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s.
v = 0 + (-9.8 m/s^2) * 5 s
v = -49 m/s
Hence, the ball's vertical velocity 5 seconds later is -49 m/s.
To determine the ball's horizontal velocity 5 seconds later, we can assume that there is no horizontal acceleration acting on the ball. This means that the ball's horizontal velocity remains constant at 10 m/s throughout its motion. Therefore, the ball's horizontal velocity 5 seconds later will also be 10 m/s.
Now let's calculate the ball's vertical velocity 5 seconds later. To do this, we need to consider the effect of gravity on the ball's vertical motion. Assuming no air resistance, the only force acting on the ball vertically is gravity, which causes it to accelerate downward at a rate of approximately 9.8 m/s^2.
After 5 seconds of free fall, the ball's vertical velocity can be calculated using the equation:
Final vertical velocity = Initial vertical velocity + (acceleration due to gravity * time)
Since the ball is thrown horizontally, it has no initial vertical velocity. Therefore, we can substitute 0 for the initial vertical velocity in the equation.
Final vertical velocity = 0 + (9.8 m/s^2 * 5 s) = 49 m/s
So, the ball's vertical velocity 5 seconds later is 49 m/s downward.