Question

A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock doesn’t hit the building on its way back down and lands on the street below. Ignore air resistance. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street?

Answer 1

Answer:

a) $v=32.74\frac{m}{s}$.

b) $t=\frac{v-v_0}{g} =5.58s$.

Explanation:

a) What is the speed of the rock just before it hits the street?

From Kinematics we have $v^2= v_0^2+2a(y_f-y_i)$.

If we take the top of the roof as the position $y_0=0m$, then

$y_f=-30m$ and we have (also, $a=g=-9.8\frac{m}{s^2}$) :

$v^2= v_0^2+ 2ay_f$ ⇒ $v= \sqrt{v_0^2+ 2ay_f}$

⇒ $v=32.74\frac{m}{s}$.

b) How much time elapses from when the rock is thrown until it hits the street?

From Kinematics we have $v(t)=v_0+at=v_0+gt$

When the rock touches the ground:

$v=-32.74\frac{m}{s}$

With a minus sign to indicate the vector velocity points down.

$t=\frac{v-v_0}{g} =5.58s$

(remember that $g=-9.8\frac{m}{s^2}$)

Answer 2

(a) Speed is defined as the rate of change of the distance or the height attained. The speed of rock when it just hit the street will be 32.75 m / sec.

(b) Elapsed time is defined as the time taken by the body to travel the given distance. The time elapsed by the rock hits the street will be 5.58 seconds.

What is speed?

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while u for the final speed. its si unit is m/sec.

(a)  Given

Initial speed of throw(u) = 22.0 m/sec

Final speed of throw against gravity(v)= ? m/sec

Initial  speed of throw against gravity(\u)=  22.0 m/sec

The height attained by the body from the point of throw(h) = 30.0 m

According to newtons third law of motion

$\rm {v^{2} = u^{2} + 2gh}$

$\rm { v^{2} = 2gh}$

$\rm {v = \sqrt{2gh} }$

$\rm {v = \sqrt{2 \times9.81 \times 30.0 } }$

$\rm {v = 32.74 m/sec.}$

Hence the speed of rock when it just hit the street will be 32.75 m/sec.

(b)

According to newtons first law of motion

Given,

v = u + at

The total time elapsed will be the sum of time taken to reach the highest point from the ground and the time taken by the body to travel to the highest point and the ground.

$t = t_1 + t_2$

$t_1 = \frac{v-u}{g}$

$t_1 = \frac{0-22}{-9.81}$

$t_1 = 2.24 sec$

$t_2 = \frac{v-u}{g}$

$\rm{t_2 = 3.33 sec }$

$t = t_1 + t_2$

$\rm{t = 2.24 + 3.33\\\\t= 5.57 sec}$

Hence the time elapsed by the rock hits the street will be 5.58 seconds.

To know more about the speed refer to the link

https://brainly.com/question/7359669