8.) A ball is thrown vertically upwards from the top of
a building of height 24.5 m with an initial velocity
19-6 m s-1. Taking g = 9.8 m s?, 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
Answer:t is stated that,
Initial velocity(u) = 19.6 m/s (upward)
Height of the building(x) = 24.5 m
So, now come to the point of calculation__
(i) The final velocity(v) = 0 m/s (at the highest point)
For upward journey, from relation,
v²= u²- 2gh
⇒ 0 = u² - 2gh
⇒ h = (u²/2g)
⇒ h = (19.6)²/2×9.8
⇒ h = 19.6 m
∴ The maximum height reached by the ball is 19.6 m.
(ii) While returning from the highest point,
u=0
total height travelled = 19.6+24.5 m
= 44.1 m
Let, v be the velocity with which it strikes the ground.
Then from relation we get,
v² = u²+2gh
∴ v² = 0+ 2×9.8×44.1
⇒ v = √2×9.8×44.1
⇒ v = 29.4 m/s
∴ The velocity with which the ball will strike the ground is 29.4 m/s
(iii) Let, the time taken by the ball to reach to the highest point be t1 and the time taken to strike the ground be t2
From the relation of upward journey we get,
v= u-gt
where, u = 0 and v = 29.4 m/s
∴ 29.4 = 0+9.8 t2
⇒ t2 = 29.4/9.8
⇒ t2 = 3 s
∴ The total time of the journey t = t1+t2
= 2+3
= 5 s
Explanation: please mark me as the brainlest