Question

A stone is thrown vertically upward with a speed of 49 m/s.Then the velocity of the stone one second before it reaches the maximum height

Answer 1

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:

$v_f = v_i + at$

0 = 49 - 9.8 × t

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

t = 5 seconds

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

Hence, time = 4 seconds

$v_f = v_i + at$

$v_f = 49 - 9.8 \times 4$

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

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

Answer 2

Answer:

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