A stone is thrown vertically upward with a velocity of 4.9 metre per second calculate the maximum height reached time taken to reach the maximum height velocity with which it returns to the ground and time taken to reach the ground
Answers
We know initial velocity (u) of the ball is 4.9 m/s
The acceleration on the ball is due to gravity = -10 m/s²
Final velocity at maximum height is 0 m/s
∴ Upon using the kinematic equation v² - u² = 2as (where s is max height reached by the ball)
= (0)² - (4.9)² = 2(-10)s
= -24 = -20×s
∴ s = 24/20 = 6/5 = 1.2 m is the max height reached.
The time taken to reach the max height can be found by using v = u+at
again substituting the above values:
0 = 4.9 -10t (∵ g = -10m/s²)
∴ t = 4.9/10 = 0.49 s is the time taken to reach the max height.
Now the ball falls from the top.
∴ The initial velocity will be 0 m/s and the acceleration is again -10m/s²
The final velocity before it hits the ground:
v = u + at (Note t will be the same as time from top to ground is same)
v = 0 -10 × 0.49
v = -4.9 m/s is velocity with wich it returns to ground (It is negative as it is falling down and is towards negative y-axis)
time taken to reach the ground is same as it took to reach up i.e 0.49 s
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