Question

11. A ball is thrown vertically upwards with an initial velocity of 47 m s-1

. Calculate: (i) The maximum

height attained, (ii) The time taken by it before it reaches the ground again. (Take g = 9.8 m s-2).
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Answer 1

Answer:

Explanation:

given,

initial velocity u=47 m/s

so, we know that final velocity in vertically upward =0

maximum height attained =h

v^2-u^2=2gh

0-(47)^2=2x9.8xh

h=2209/19.6

h=112.7m

the time taken to reach the ground again =t

v=u-gt

0=47-9.8*t

t=4.79sec

Answer 2

Given :

• Initial velocity, u = 47 m/s
• Final velocity, v = 0 m/s
• Acceleration due to gravity, g = - 9.8 m/s²

To find :

• Height attained, h
• Time taken, t

According to the question,

➞ v² = u² + 2gh

Where,

• v = Final velocity
• u = Initial velocity
• g = Acceleration due to gravity
• h = Height

➞ (0)² = (47)² + 2 × (-9.8)× h

➞ 0 = 2209 - 19.6h

➞ 0 - 2209 = - 19.6h

➞ - 2209 = - 19.6h

➞ 2209 ÷ 19.6 = h

➞ 112.7 = h

So,the height attained is 112.7 meter.

Now,

➞ v = u + gt

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time

➞ 0 = 47 + (-9.8) × t

➞ 0 - 47 = - 9.8t

➞ - 47 = - 9.8t

➞ 4.79 = t

So,the time taken is 4.79 seconds.