All categories

3 years ago

A boy throws a ball vertically up and catches it after 3 s. What height did the ball reach?

Physics

2 answers:

Crazy boy [7]3 years ago

7 0

Answer:

11.025 meters

Explanation:

The gravity’s acceleration on the surface of earth is approximately 9.8m/s

If it took 3 s to go up and come back, that means it took (1/2) × 3 to go from his hand to its highest point, and that amount of time again (1.5 seconds) to go back down to his hand.

This means the ball accelerated from Vi (initial Velocity) to 0 in 1.5 s

To find g in 1.5 i.e speed =

1.5 × 9.8 m/s = 14.7 m/s.

To find the height the ball reached=

We will use d = rt for this. Distance = Rate × Time

Since, it went 14.7 m/s at the start, and 0 m/s at the top, so the average speed of the ball going upward was (0+14.7)/2 = 7.35 m/s

7.35 m/s was the average speed = rate.

Time was 1.5 seconds, so we multiply.

7.35 m/s × 1.5seconds = 11.025 meters

Darya [45]3 years ago

3 0

The top height the ball had is (9/8)*g
g being the strength of the earth's attraction. so if g is about 9.8 then the ball went up about 11 meters

You might be interested in

An unbanked circular highway curve on level ground makes aturn

tino4ka555 [31]

Answer:

Explanation:

Given

Velocity of traffic $v=60\ mi/hr\approx 26.82\ m/s$

Maximum value of Centripetal force is one-tenth of weight

such that $(F_c)_{max}=\frac{mg}{10}$

to make a safe turn centripetal force with radius of curvature r is given by

$F_c=\frac{mv^2}{r}$

$(F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}$

$\frac{mg}{10}=\frac{mv^2}{r}$

$r=\frac{10v^2}{g}$

$r=\frac{10\times 26.82^2}{9.8}$

$r=733.99\approx 734\ m$

4 0

4 years ago

What happens inside someone's body when they danve

LUCKY_DIMON [66]

The body releases endorphins

6 0

3 years ago

As more and more bulbs are connected in series to a flashlight battery, what happens to the brightness of each bulb? Assuming th

balu736 [363]

Answer:

$P_1 = P_2 = \frac{P}{2}$

so each bulb brightness becomes half of its given or indicated power

$P_1 = P_2 = \frac{V^2}{R}$

so both bulb will glow same power as indicated

Explanation:

Let the indicated power on the bulbs is given as P and its rated voltage is V

so here resistance of each bulb is given as

$R = \frac{V^2}{P}$

now if the two bulbs are connected in series so we will have

$R_{eq} = R_1 + R_2$

$R_{eq} = 2\frac{V^2}{P}$

now the current in the circuit is given as

$i = \frac{V}{R_{eq}$

$i = \frac{P}{2V}$

now brightness of each bulb is given as

$P_1 = P_2 = i^2 R$

$P_1 = P_2 = \frac{P}{2}$

so each bulb brightness becomes half of its given or indicated power

Now if the two bulbs are connected in parallel

then the net voltage across each bulb is "V"

so we will have

$P_1 = P_2 = \frac{V^2}{R}$

so both bulb will glow same power as indicated

6 0

4 years ago

A uniform brick of length 25 m is placed over

Harlamova29_29 [7]

The system of two rods will lie on the table as show in the figure

The center of first rod will lie exactly at the edge of first rod and then the center of mass of two combined rods will lie at the edge of the table.

So now the whole system will rest on the rod and it will not tipping off.

Since both rods are identical so we can say that the system will have its center of mass at the mid point on the line joining the two centers

So the value of x will be at mid point of line joining the two points on rod

$x = \frac{12.5}{2} = 6.25 cm$

So total length over the edge will be given as

$L = 6.25 cm + 12.5 cm = 18.75 cm$

3 0

3 years ago

Karolina [17]

<h2>Carry energy</h2>

- - - - - - - - - - - - - - - - - - - - - - -

Hope this helped :)

~pinetree

5 0

2 years ago