A stone is thrown vertically upwards with a velocity of 9.8 ms–1. When will it reach the ground?
Answers
Answer:
We know the Initial Velocity (V0), the Acceleration (around 9.8m/s2, since it’s gravity). We need to figure out the time for both ascending and descending, and also the distance between the ground and it’s maximum point (where it’s velocity will be 0).
Explanation:
The equation we will need is one of the Kinematic equations:
Vf=V0+at
This is used to find the final velocity of an object under acceleration, after a certain time. If we know the final velocity, but not the time, we can still use it to solve for t.
Let’s divide the trip of the ball in two: when it goes up, and when it comes down. We’ll do that because the time will be the same in both cases (ignoring air resistance). Let’s just concentrate on up, because it is the only one with which we know enough to solve the rest.
When the ball reaches it’s highest point, it will be at zero velocity, preparing to come down. So if we know that V0=9.8m/s, a=−9.8m/s2 and that Vf=0m/s, then we can easily find the time it takes by plugging everything into the equation. Note: The negative sign on the Acceleration is because when the ball goes up, gravity slows it down, since it’s direction is opposite to the direction of the ball.
Now let’s find t:
Vf=V0+at
0m/s=9.8m/s+(−9.8m/s2∗t)
−9.8m/s=−9.8m/s2∗t
−9.8m/s−9.8m/s2=t
t=1s
The time it will take for the ball to reach it’s maximum height is 1 second. Now, we just need to double this to get the whole trip of the ball, up and down, because it takes the same time. So the answer is that the ball, if thrown from the ground will take 2 seconds to come back down.
Answer:
Let’s first write everything we know, and then we can start working with the equation that we need to figure out the time.
We know the Initial Velocity (V0), the Acceleration (around 9.8m/s2, since it’s gravity). We need to figure out the time for both ascending and descending, and also the distance between the ground and it’s maximum point (where it’s velocity will be 0).
The equation we will need is one of the Kinematic equations:
Vf=V0+at
This is used to find the final velocity of an object under acceleration, after a certain time. If we know the final velocity, but not the time, we can still use it to solve for t.
Let’s divide the trip of the ball in two: when it goes up, and when it comes down. We’ll do that because the time will be the same in both cases (ignoring air resistance). Let’s just concentrate on up, because it is the only one with which we know enough to solve the rest.
When the ball reaches it’s highest point, it will be at zero velocity, preparing to come down. So if we know that V0=9.8m/s, a=−9.8m/s2 and that Vf=0m/s, then we can easily find the time it takes by plugging everything into the equation. Note: The negative sign on the Acceleration is because when the ball goes up, gravity slows it down, since it’s direction is opposite to the direction of the ball.
Now let’s find t:
Vf=V0+at
0m/s=9.8m/s+(−9.8m/s2∗t)
−9.8m/s=−9.8m/s2∗t
−9.8m/s−9.8m/s2=t
t=1s
The time it will take for the ball to reach it’s maximum height is 1 second. Now, we just need to double this to get the whole trip of the ball, up and down, because it takes the same time. So the answer is that the ball, if thrown from the ground will take 2 seconds to come back down.