Question

If a ball is dropped off a tall building and accelerates at 9.8 m/s^2 until reaching the ground at a speed of 55 m/s, how long will it take for the ball to reach the ground?

Answer 1

5.61s

Step-by-step explanation:

`\pink{\frak{Given}}\begin{cases} \textsf{ A ball is dropped off a tall building and accelerates at 9.8 m/s$^2$ .}\\\textsf{It reaches the ground at a speed of 55 m/s .} \end{cases}`

And we need to find out the time taken by the ball to reach the ground .

• The initial velocity of the ball will be 0 since it is dropped .

Here we can use the First equation of motion , namely ;

`\sf \longrightarrow v = u + at`

where the symbols have their usual meaning . Now substituting the respective values , we have,

`\sf \longrightarrow 55m/s = 0m/s + 9.8 m/s^2(t)\\`

`\sf \longrightarrow 55m/s = 9.8m/s^2(t)\\`

`\sf \longrightarrow t = (55m/s)/(9.8m/s^2)\\`

`\sf \longrightarrow \boxed{\bf time = 5.61 s }`