A ball is thrown vertically upwards from the top of a building of height 24.5 m with initial velocity 19.6 m/s. Taking g=9.8 m/s², calculate:(i) the height to which it will rise before returning to the ground, (ii) the velocity with which it will strike the ground and (iii) the total time of journey.
Answers
Given
height of building = 24.5m
initial velocity = 19.6m/s
g = 9.8m/s^2
At maximum hight the value of v =0
so
v^2 = u^2 - 2gh(max )
0 = (19.6)^2 - 2*9.8 * h(max)
2*9.8*h(max) = (19.6)^2
h(max)= 19.6*19.6/2*9.8 = 19.6m
Now
initial velo velocity of body = 0
h(max ) = 19.6m+ 25.4m (building hight)= 45m
g = 9.8m/s^2
Now
u^2 - 2h(max )g = v^2
- 2 *45*9.8 = v^2
-882 = v^2
v = - 21sqrt2m/s
v= 21sqrt2m/s towards gravity
time taken When body thrown
vertically upward
Hmax = ut - gt^2/2
19.6 = 19.6t - 19.6t^2/2
t^2 - t - 1/2 = 0
t^2 - 2t + 1/4 - 3/4 = 0
(t - 1/2)^2 - (sqrt3/2)^2 = 0
t = (1 + sqrt3)/2s and (1 - sqrt3)/2 s
t = 1.37 s
time taken in free fall body
hmax = ut - gt^2/2
45 = 19.6t^2/2
t^2 = 90/9.6
t^2 = 9.375
t = 3.06s
so total time = 1.37 + 3.06 = 4.43s
(i) The height to which it will rise before returning to the ground = 19.6 m
(ii) The velocity with which it will strike the ground is = 29.4 m/s.
(iii) The total time of journey is = 5 s
Given: initial velocity (u) = 19.6 m/s, height of building (h) = 24.5 m
To Find:
(i) the height to which it will rise before returning to the ground.
(ii) the velocity with which it will strike the ground.
(iii) the total time of journey.
Solution:
(i) initial velocity (u) = 19.6 m/s
height of building (h) = 24.5 m
We know that at the highest point velocity becomes zero.
So, final velocity (v) = 0
Applying the formula,
v² = u² - 2gh [ negative is used since velocity is decreasing ]
⇒ 0 = 19.6² - 2 × 9.8 × h
⇒ h = ( 19.6 )² / ( 2 × 9.8 )
= 19.6 m
The height to which it will rise before returning to the ground = 19.6 m
(ii) For the second phase of the journey where the ball strikes the ground,
the initial velocity (u) = 0,
the total distance travelled = 19.6 + 24.5 m
= 44.1 m
Let the velocity with which it will strike the ground be 'v'.
Applying the formula,
v² = u² - 2gh
where u = 0, g = 9.8 m/s² , h = 44.1 m. Putting these values in (1), we get;
⇒ v² = 0 + 2 × 9.8 × 44.1
⇒ v² = 864.36
⇒ v = √( 864.36 )
⇒ v = 29.4 m/s
The velocity with which it will strike the ground is 29.4 m/s.
(iii) Let the time taken by the ball to reach the highest point be ' t1 ' and the time taken by the ball to reach the ground be ' t2 '.
For calculating t1, we need to use the formula,
v = u - gt [ negative is used since velocity is decreasing ]
Here, v = 0 , u = 19.6 m/s, g = 9.8 m/s²
Putting the values in the formula we get,
⇒ 0 = 19.6 - 9.8 × t1
⇒ t1 = 19.6 / 9.8
⇒ t1 = 2 s
For calculating t2, we need to use the formula,
v = u + gt [ since ball is coming down so its velocity is increasing ]
Here, v = 29.4 m/s , u = 0 , g = 9.8 m/s²,
Putting the values in the formula we get,
⇒ 29.4 = 0 + 9.8 × t2
⇒ t2 = 29.4 / 9.8
⇒ t2 = 3 s
Now, the total time of journey is = t1 + t2 = ( 2 + 3 ) s
= 5 s
Hence, compiling all the answers we get,
(i) The height to which it will rise before returning to the ground = 19.6 m
(ii) The velocity with which it will strike the ground is = 29.4 m/s.
(iii) The total time of journey is = 5 s
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