Question

A stone is thrown vertically upward with a speed of 49 ms-!. Then the velocity of the stone, one second
before it reaches the maximum height is
​

Answer 1

Answer:

No

Explanation:

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Answer 2

Explanation:

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Initial velocity (u) = 49 m/s

Let the time taken to reach the maximum height be t.

Acceleration due to gravity = - 9.8 m/s^2

Final velocity = 0 m/s

Now, using first equation of motion:

v_f = v_i + atv

f

=v

i

+at

0 = 49 - 9.8 × t

t = \frac{49}{9.8}t=

9.8

49

t = 5 seconds

Now, we need to find the velocity just 1 second before reaching the highest point.

Hence, time = 4 seconds

v_f = v_i + atv

f

=v

i

+at

v_f = 49 - 9.8 \times 4v

f

=49−9.8×4

v_f = 9.8 \frac{m}{s}v

f

=9.8

s

m

Hence, the velocity just 1 second before reaching the highest point is 9.8 m/s.