A BALL IS THROWN VERTICALLY UPWARDS FROM GROUND LEVEL HIT THE GROUND AFTER 4 SECONDS CALCULATE THE MAXIMUM HEIGHT IT REACH DURING THE JOURNEY(g=10m|s)
max height reached at 2 seconds. use that.
unknown
To calculate the maximum height the ball reaches during its journey, we can use the equations of motion for vertical motion with constant acceleration.
Given:
Time taken for the ball to hit the ground, t = 4 seconds
Acceleration due to gravity, g = 10 m/s^2
We know that when the ball reaches its maximum height, its final velocity is zero. Using the equation:
vf = vi + gt
Where:
vf = final velocity
vi = initial velocity
g = acceleration due to gravity
t = time taken
In this case, the final velocity is zero when the ball reaches its maximum height. So, we can rewrite the equation as:
0 = vi + gt
Simplifying the equation, we find:
vi = -gt
The initial velocity can be determined by using the equation for vertical displacement:
h = vi*t + (1/2)gt^2
Where:
h = maximum height reached
Substituting the value of vi from the equation above:
h = (-gt)*t + (1/2)gt^2
Simplifying this equation, we get:
h = -(1/2)gt^2
Substitute the given values:
h = -(1/2)(10)(4^2)
h = -(1/2)(10)(16)
h = -(1/2)(160)
h = -80 meters
Since the maximum height cannot be negative, we can take the magnitude of the result:
Maximum height = 80 meters
Therefore, the maximum height the ball reaches during its journey is 80 meters.
To calculate the maximum height the ball reaches during its journey, we can use the equation of motion for vertically thrown objects:
h = u*t + (1/2)*a*t^2
Where:
h = maximum height
u = initial velocity
t = time taken to reach maximum height
a = acceleration due to gravity
Given:
u = unknown
t = 4 seconds
a = -10 m/s^2 (negative as it is acting in the opposite direction to the ball's motion when it reaches maximum height)
Since the ball's velocity is zero at the maximum height, we can set the final velocity (v) equal to zero and use the formula:
v = u + a*t
0 = u - 10*4
u = 10*4
u = 40 m/s
Now that we have the initial velocity, we can substitute it into the equation to find the maximum height:
h = u*t + (1/2)*a*t^2
h = 40*4 + (1/2)*(-10)*(4^2)
h = 160 - 80
h = 80 meters
Therefore, the maximum height the ball reaches during its journey is 80 meters.