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
v=0
g= -9.8 m/s
2gh=v^2-u^2
2*-9.8*h=-2401
h=122.5 m
v=u+gt
0=49+(-9.8)t
-49=-9.8t
t=5 sec
So Total time=5+5=10 sec
Given:
The initial velocity u = 49 m/s
To Find:
(i) The maximum height to which it rises
(ii) The total time it takes to return to the surface
Solution:
The maximum height of the ball is 122.5 m and the total time taken by the ball to return to the surface is 10s.
i) At the highest point the velocity of the ball v = 0 m/s
During upward motion, g = − 9.8 m s²
Let h be the maximum height attained by the ball.
Since the acceleration is constant, we can use the 3rd equation of motion.
v² - u² = 2 a.s
Substituting the values,
- 49² = 2 X (-9.8) h
or h = 49 X 49 / 2 X 9.8
⇒ h = 122.5 m
ii) Let t be the time taken by the ball to reach the maximum height h.
Using 1st equation of motion,
v = u + (-g)t
Substituting,
49 = 9.8 X t
or 9.8 t = 49
or t= 49/9.8
= 5 s
Since we know that the time of ascent = time of descent
⇒ The total time taken by the ball to return to the surface = 2 X t
2 X 5
= 10s