A ball thrown vertically upwards from ground level hits the ground after 4s. Calculate the maximum height it reached during during it journey (g=10m/s2)
time up = time down = t
h = 1/2 g t^2
To calculate the maximum height reached by the ball, we can use the equation for motion under constant acceleration:
h = (v^2 - u^2) / (2g)
where:
h = maximum height
v = final velocity
u = initial velocity
g = acceleration due to gravity
In this case, we know that the ball hits the ground after 4 seconds, which means its total time of flight is 8 seconds (the time it takes to go up and come back down).
Since the ball is thrown vertically upwards, its final velocity when it hits the ground will be -u.
Using the equation v = u + gt, we can find the initial velocity (u) of the ball:
- u = u + gt
Rearranging the equation, we have:
- u - u = gt
0 = gt
Therefore, u = 0 m/s.
Now we can substitute the values into the equation for the maximum height:
h = (v^2 - u^2) / (2g)
h = (-u^2) / (2g)
h = 0^2 / (2 * 10)
h = 0 meters
Therefore, the maximum height reached by the ball during its journey is 0 meters.
To calculate the maximum height reached by the ball, we can use the equations of motion. The key equation in this case is the one that relates the final velocity of the ball (when it hits the ground) to its initial velocity and the acceleration due to gravity. This equation is:
v^2 = u^2 + 2as
where:
v is the final velocity (which is 0 when the ball hits the ground),
u is the initial velocity (unknown in this case, as we are trying to find the maximum height),
a is the acceleration due to gravity (which is -g, where g is the acceleration due to gravity and its value is 10 m/s^2 in this case),
and s is the displacement (also unknown, as we are trying to find the maximum height).
In this scenario, since the ball is thrown upwards, its initial velocity will be positive. When the ball hits the ground, its final velocity is 0. Therefore, we can rewrite the equation as:
0 = u^2 + 2as
We can solve this equation to find the displacement (maximum height reached):
0 - u^2 = 2as
s = -u^2 / (2a)
Now, we know that the time it takes for the ball to reach its maximum height is half of the total time it takes for the ball to hit the ground. So the time taken to reach the maximum height is 4s / 2 = 2s.
Using another equation of motion, we can relate the displacement, initial velocity, and time:
s = ut + (1/2)at^2
Since the ball is at the maximum height, its displacement is the maximum height, and we can rearrange the equation:
s = ut - (1/2)gt^2
Substituting the values we know:
s = ut - (1/2)gt^2
s = u * 2 - (1/2) * 10 * (2^2)
s = 2u - 20
Now, we can substitute this value of s in the equation we derived earlier:
2u - 20 = -u^2 / (2a)
2u - 20 = -u^2 / (-20)
2u - 20 = u^2 / 20
Multiplying both sides by 20:
40u - 400 = u^2
Rearranging the equation:
u^2 - 40u + 400 = 0
Now we can solve this quadratic equation to find the value of u.
Using the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / 2a
where a = 1, b = -40, and c = 400.
Calculating the values:
u = (-(-40) ± √((-40)^2 - 4 * 1 * 400)) / (2 * 1)
u = (40 ± √(1600 - 1600)) / 2
u = (40 ± √0) / 2
u = (40 ± 0) / 2
u = 40 / 2
u = 20
Therefore, the initial velocity (u) of the ball is 20 m/s.
Now, substituting this value into the equation s = 2u - 20:
s = 2 * 20 - 20
s = 40 - 20
s = 20
Hence, the maximum height reached by the ball during its journey is 20 meters.