Please help me to solve this problem.
A body is projected vertically upwards with a velocity 20m/s. find the maximum height reached by the body.(g=10m/sec2)
Let initiql velocity be 10^2 and final velocity be 0 and acceletation due to gravity be -10m/s^2
By using formula
V^2- u^2=2*a*S
0^2- 10*2 =2*-10S
-100=-20S
-100/-20=s
5m=s
Enough already !
Let initiql velocity be 20^2 and final velocity be 0 and acceletation due to gravity be -10m/s^2
By using formula
V^2- u^2=2*a*S
0^2- 20^2=2*-10S
-400=-20S
-400/-20=s
20m=s
Let initial velocity be 20^2 and final velocity be 0 and acceletation due to gravity be -10m/s^2
By using formula
V^2- u^2=2*a*S
0^2- 20^2=2*-10S
-400=-20S
-400/-20=s
20m=s
H=u^2/2g
=20×20/2×10
=20m/s
To find the maximum height reached by the body, we can use the kinematic equation for displacement in vertical motion.
The equation is:
s = u⋅t - 0.5⋅g⋅t²
Where:
s = displacement (maximum height)
u = initial vertical velocity = 20 m/s (upwards)
g = acceleration due to gravity = 10 m/s² (negative as it acts downwards)
t = time taken to reach maximum height
We need to find the value of t when the body reaches its maximum height. At this point, the vertical velocity becomes zero (since the body momentarily stops moving upwards before starting to fall back down).
Using the formula for final velocity in vertical motion, we have:
v = u - g⋅t
Since v = 0 (at maximum height), we can solve for t:
0 = 20 - 10⋅t
10⋅t = 20
t = 20/10
t = 2 seconds
Now that we have the value of t, we can substitute it back into the displacement equation to find the maximum height:
s = u⋅t - 0.5⋅g⋅t²
s = 20⋅2 - 0.5⋅10⋅2²
s = 40 - 0.5⋅10⋅4
s = 40 - 0.5⋅40
s = 40 - 20
s = 20 meters
Therefore, the maximum height reached by the body is 20 meters.
usquare divided by 2g
therefore 400/20 is 20 metres
thankyou