A ball is thrown vertically upwards.It takes 4 s for the ball to return to the thrower's hand now what is thb maximum height reached by the ball.
you need more information to solve the problem, such as the initial or final velocity.
If you have this information you can solve for d (displacement) using the equation (s):
- Vf^2= Vi^2 + 2*g*d where g is the acceleration due to gravity (9.8 m/s^2)
- d= Vi*T+0.5*gT^2 where t is time
Hope this helps
To find the maximum height reached by the ball, we can use the formula for the displacement of an object in free fall. Since the ball is being thrown vertically upwards and then comes back down, the total displacement will be zero.
The formula for displacement is:
Displacement = Initial Velocity * Time + (1/2) * Acceleration * Time^2
In this case, the displacement is zero, as the initial and final points are the same. We also know that the acceleration due to gravity, denoted as 'g', is acting downwards. Therefore, we can rewrite the formula as:
0 = Initial Velocity * Time - (1/2) * g * Time^2
Simplifying this equation, we get:
(1/2) * g * Time^2 = Initial Velocity * Time
Now, we need to find the initial velocity of the ball. Since it takes 4 seconds for the ball to return to the thrower's hand, we can use this information to find the initial velocity.
From the given information, the time taken to reach the maximum height is half the total time taken for the complete journey. So, the time taken to reach the maximum height is 4s/2 = 2s.
Now, we can plug in the values into the equation to solve for the initial velocity:
(1/2) * g * (2s)^2 = Initial Velocity * 2s
Simplifying further:
2 * g * 4s^2 = Initial Velocity * 2s
8 * g * s^2 = Initial Velocity * 2s
Canceling the common factor of '2s':
4 * g * s = Initial Velocity
Now we have the initial velocity of the ball. To find the maximum height reached by the ball, we need to find the displacement at half of the total time. Since it is a symmetrical motion, the maximum height is reached halfway through the total time taken.
So, plugging in the values into the displacement formula:
Displacement = Initial Velocity * Time + (1/2) * g * Time^2
Displacement = (4 * g * s) * (4s/2) + (1/2) * g * (4s/2)^2
Displacement = 8 * g * s^2 + g * s^2
Displacement = 9 * g * s^2
We know that the displacement at the maximum height is equal to the height reached by the ball. Therefore, the maximum height reached by the ball is 9 times the acceleration due to gravity (g) multiplied by the square of the time taken to reach the maximum height (s).