A ball thrown vertically upward and reached to thrower in 6sec what will be velocity and man maximum hight
To find the velocity and maximum height of a ball thrown vertically upward, we can use the equations of motion.
First, let's identify what information we have:
Time taken (t) = 6 seconds
We can use the equation for displacement:
s = ut + (1/2)at^2
Since the ball is thrown upward, the initial velocity (u) will be positive. We can assume that the acceleration (a) due to gravity is -9.8 m/s^2 (negative because it opposes the upward motion).
At the maximum height, the final velocity (v) will be zero, so we can use the equation:
v = u + at
Now, let's find the velocity and maximum height step by step:
Step 1: Finding the initial velocity (u) using equation 1.
We know that at the highest point, the vertical displacement (s) is zero. Therefore, we can rewrite equation 1 as:
0 = u(6) + (1/2)(-9.8)(6)^2
Simplifying the equation:
-588 = 36u
Dividing both sides by 36:
u = -588/36
u ≈ -16.33 m/s
So, the initial velocity of the ball is approximately -16.33 m/s.
Step 2: Finding the maximum height (h) using equation 2.
We know that at the highest point, the velocity (v) is zero. Therefore, we can use equation 2:
0 = -16.33 + (-9.8)t
Simplifying the equation:
16.33 = 9.8t
Dividing both sides by 9.8:
t ≈ 1.67 s
So, it takes approximately 1.67 seconds for the ball to reach its highest point.
Step 3: Calculating the maximum height (h) using equation 1.
Now that we know the time taken to reach the highest point, we can substitute this value into equation 1:
h = (-16.33)(1.67) + (1/2)(-9.8)(1.67)^2
Simplifying the equation:
h ≈ -27.27 m
The negative sign indicates that the height is measured below the starting point. Taking the magnitude, the maximum height reached by the ball is approximately 27.27 meters.