Question

A stone is thrown vertically upward with a velocity of 98m/s. What will be its maximum height attend and time taken to achive this height?

Answer 1

To find the maximum height attained by the stone and the time taken to achieve this height, we can use the equations of motion for vertical motion under constant acceleration.

Given: Initial velocity (v₀) = 98 m/s (upward) Acceleration due to gravity (a) = -9.8 m/s² (downward)

1. Maximum Height: At the maximum height, the stone will momentarily come to rest, so its final velocity (vf) will be 0 m/s. We can use the equation:

vf = v₀ + a * t

0 = 98 - 9.8 * t

Solving for time (t), we find:

t = v₀ / a

t = 98 / 9.8

t ≈ 10 s

Now, to find the maximum height (h), we can use the equation:

h = v₀ * t + (1/2) * a * t²

h = 98 * 10 + (1/2) * (-9.8) * (10)²

h = 980 - 490

h = 490 meters

Therefore, the stone reaches a maximum height of 490 meters.

2. Time taken to achieve maximum height: As calculated earlier, the time taken to reach the maximum height is approximately 10 seconds.

Thus, the stone takes approximately 10 seconds to achieve its maximum height.