if a ball is thrown vertically into the air, leaving the ground at a height of 1.8m, with a speed of 9.7m s -1. When it rises and falls to the ground what will the balls speed be as it hits the ground? please help with an explanation as i am so stuck, my exam is in two days and i need help.
thanx
Becky
I would do this with energy.
Final total energy=initial total energy
1/2 m vf^2+zeroPE=mg(hi)+1/2 mvi^2
Vf^2=2g*1+vi^2
you may have had this formula presented to you for memory as..
Vf^2=Vi^2+2ad
Thank you Bob Pursley :)
Penejoldo
To find the speed of the ball as it hits the ground, we need to analyze the motion of the ball.
First, let's consider the ball at its maximum height, where its vertical velocity is zero. From this point, the ball will start falling back towards the ground.
The key thing to note here is that when the ball reaches the ground, it will have the same speed as it had when it was initially thrown upwards. This is because the force of gravity only affects the ball's vertical motion, not its horizontal speed.
Now, to find the speed of the ball when it was initially thrown upwards, we can use the equation for final velocity (v) in terms of initial velocity (u), acceleration (a), and time (t):
v = u + at
In this case, the ball is thrown vertically, so the acceleration is due to gravity and is equal to -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).
Given that the initial velocity (u) is 9.7 m/s, and we want to find the final velocity (v) when the ball hits the ground, we can rearrange the equation:
v = u + at
Since the ball is thrown upwards, the final velocity (v) would be negative when it returns to the ground.
Plugging in the values:
v = 9.7 m/s + (-9.8 m/s^2) * t
We need the time (t) it takes for the ball to reach the maximum height and fall back to the ground. The time it takes for the ball to reach maximum height is the time it takes for the vertical velocity to become zero. Using the equation:
v = u + at
0 = 9.7 m/s + (-9.8 m/s^2) * t_max
Simplifying:
9.8 m/s^2 * t_max = 9.7 m/s
t_max = 9.7 m/s / 9.8 m/s^2
t_max ≈ 0.99 seconds
To find the total time for the ball to reach the ground, we know that the time it takes to reach maximum height is the same as the time it takes to fall back down to the ground. Therefore:
t_total = 2 * t_max
t_total ≈ 2 * 0.99 seconds
t_total ≈ 1.98 seconds
Now, we can plug the total time back into our original equation to find the final velocity (v) when the ball hits the ground:
v = 9.7 m/s + (-9.8 m/s^2) * t_total
v = 9.7 m/s + (-9.8 m/s^2) * 1.98 seconds
v ≈ 9.7 m/s - 19.4 m/s
v ≈ -9.7 m/s
Therefore, the speed of the ball as it hits the ground is approximately 9.7 m/s in the downward direction.
Remember to double-check the calculations and units to ensure accuracy. Good luck on your exam!