Question

A ball is thrown vertically with a velocity of 49 m/s. Calculate:
i) the maximum height to which it rises,
ii) the total time it takes to return to the surface of the earth.​

Answer 1

Answer:

U=49 m/s

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

Total time=5+5=10 sec.

Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\bold\purple{Answer:-}$

⏩ i) In Cartesian sign convention, upward velocity is taken positive and acceleration due to gravity is taken negative.

Therefore,

u= + 49 m/s, g= -9.8 ms^-2,

At the highest point, v= 0

Now, as v² - u² = 2gs,

Therefore,

0² - 49² = 2 (-9.8) × s

So, maximum height, s= 49 × 49/2 × 9.8

= 122.5 m. Ans.

ii) Let t be the time taken by the stone to reach the highest point.

As, v= u + gt,

Therefore,

0 = 49 - 9.8 × t

→ t = 49/9.8

→t= 5 s

Since,

Time of ascent = Time of descent,

Therefore,

Time taken by the stone to return to the earth's surface

= 2t = 2 × 5 = 10 s. Ans.

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