A ball is thrown vertically upwards from ground level hits the ground after 4secs, calculate the maximum hieght it reached during its journey.
time up equals time down
h = 1/2 g t^2 =
1/2 * 32 ft/s^2v * (2 s)^2
Don't understand
To calculate the maximum height reached by the ball, we can use the kinematic equation for vertical motion:
y = y0 + v0*t - (1/2)*g*t^2
where:
y = height (in this case, the maximum height reached)
y0 = initial height (which is 0 in this case, as the ball starts from ground level)
v0 = initial velocity (which is the same as the final velocity when the ball hits the ground)
t = time taken (which is the total time taken for the ball to go up and come back down, given as 4 seconds)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the ball is thrown vertically upwards, its initial velocity will be positive (upwards) and the acceleration due to gravity will be negative (downwards).
Now, let's calculate the maximum height reached:
Using the equation:
y = 0 + v0*t - (1/2)*g*t^2
Substituting the given values:
0 = v0*4 - (1/2)*9.8*(4^2)
Rearranging the equation to solve for v0:
v0 = (1/2)*9.8*(4^2)/4
v0 = 19.6 m/s
Now substituting this value of v0 back into the original equation, we can find the maximum height reached:
y = 0 + 19.6*4 - (1/2)*9.8*(4^2)
y = 78.4 - 78.4
y = 0
Therefore, the maximum height reached by the ball during its journey is 0 meters. This means that the ball goes directly upwards and returns back to the ground without reaching any higher point.