A Ball is thrown vertically upwards with a velocity of 49m/s calculate. i) The Maximum height to which it rises. ii) The total time it takes to return to the surface of the earth.
Answers
Answer:
1) The Maximum height to which the ball rises is 122.5m
2) The total time ball takes to return to the surface of the earth is 10 Sec
Step-by-step Explanation :
Given : Initial velocity (u) = 49 m/s
To find : 1) The maximum height to which it rises = ?
2) The total time it takes to return to the surface of the earth = ?
Given values,
initial velocity of the ball (u) = 49 m/s and
Final velocity of the ball (v) = 0
Acceleration due to gravity, g = - 9.8 m/s²
Let h be the maximum height attained by the ball.
According to the equation of motion under gravity,
v² - u² = 2gh
Substituting the given values in above equation, we get
0 - (49)² = 2 × (- 9.8) × h
(49)² = 19.6h
h = (49)²/ 19.6
h = 122.5m
∴ Height achieved by the ball, h = 122.5 m.
Now, let t be the time taken by the ball to reach the height 122.5 m then,
According to the equation of motion,
v = u + gt
Substituting the values in the above equation , we get
0 = 49 + (-9.8) × t
9.8t = 49
t = 49/9.8
t = 5 sec
But, Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 sec