An object is thrown in vertically
upward direction with a velocity of
10 m/s. calculate the maximum
height attained by the object. Also
find the time taken to reach?
(take a=10m/s^2).
Answers
Many of the answers here provide an equation, and simply suggest to input values and isolate for the variable. However, there is a much simpler manner to examine this question.
If a ball is thrown vertically in the air, with a velocity of 10 m/s, there is only one significant force which is acting against the ball (assuming air resistance is negligible). That force would be gravity. The acceleration due to gravity is known to be 10m/s2 .
(I took the value as 10, rather than 9.8, 9.81, or any other more specific value because it simplifies the explanation, and more significant digits are not necessarily required given the value for velocity is only one significant digit.)
What this essentially means is that the velocity of the ball in the vertical axis decreases by 10m/s, for every second that it is in motion.
In order to find the time the ball reaches the maximum height, you must find when the velocity of the ball in the vertical axis reaches 0. We could simply take the initial velocity, and dividing by the value of acceleration due to gravity.
(10m/s)/(10m/s2)=1sec
Therefore, the time the ball takes to reach the maximum height is 1 second.
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Answer:
S = 5 metres
t = 1 second
Explanation:
Given :
- Initial velocity of the object = u = 10 m/s
- Acceleration of the object = a = 10 m/s²
To find :
- Maximum height attained
- Time taken to reach the maximum height
Maximum height attained = u²/2g
Maximum height attained = 10²/2×10
Maximum height attained = 100/20
Maximum height attained = 5 metres
Now using the second equation of motion :
S=u×t+½×at²
5=10×t+½×-10×t²
5=10t-5t²
-5t²+10t-5=0
By solving the above quadratic equation, T = 1 second