Question

A ball is thrown vertically upward with a speed of 24.0 m/s.(a) How high does it rise? m(b) How long does it take to reach its highest point? s(c) How long does the ball take to hit the ground after it reaches its highest point? s(d) What is its velocity when it returns to the level from which it started? m/s

Answer 1

(a) 29.4 m

(b) 2.45 s

(c) 2.45 s

(d) -24 m/s

Explanation:

Given:

Initial speed (u) = 2 m/s

(a)

We can calculate the height using the following formula:

`\begin{gathered} h=(u^2)/(2g) \\ \\ \text{ We replacing:} \\ \\ h=(24^2)/((2)(9.8)) \\ \\ h=29.4\text{ m} \end{gathered}`

(b)

Now, we calculate the time at this height, as follows:

`\begin{gathered} t=(u)/(g) \\ \\ \text{ We replacing} \\ \\ t=(24)/(9.8) \\ \\ t=2.45\text{ s} \end{gathered}`

(c)

The time to reach the ground is the same as the time to reach its highest point, therefore the time is 2.45 seconds

(d)

The speed when it returns to the level from which it started is:

`\begin{gathered} v=-u \\ v=-24\text{ m/s} \end{gathered}`

It is negative because it is opposite direction