A flight into space by a spacecraft where the spacecraft returns to Earth without achieving orbit is called a

A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

Ede4ka [16]

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (17.9)² / (2 x 9.8)

h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

3 0

2 years ago

PLEASE HELP ME WITH THSES QUESTIONS 3 SENTENCES PLEASE I BEG YOU!

JulsSmile [24]

Question 2 is because the passengers have inertia, which is a tendency to resist change in motion

6 0

3 years ago

Objects 1 and 2 attract each other with a electrostatic force of 36.0 units. If the charge of Object 1 is doubled AND the distan

melamori03 [73]

Answer:

8 units

Explanation:

F=(k*q1*q2)/(r^2)

K is a constant, q1 is charge of 1, q2 is charge of 2, r is distance between the two.

6 0

2 years ago

Which of these is an example of a mechanical wave? sunlight light from a flashlight microwaves used to heat food police siren

harina [27]

Police siren is the correct answer. Hope this helps.

6 0

2 years ago

Read 2 more answers

A diver running 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. how hi

Elodia [21]

<u>Answer:</u>

a) Height of cliff = 44.145 meter

b) 7.5 meter far from its base the diver hit the water.

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Diver's vertical motion:

Initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate displacement when time is 3 seconds.

$s= ut+\frac{1}{2} at^2\\ \\ =0*3+\frac{1}{2} *9.81*3^2\\ \\ =44.145meter$

So height of cliff = 44.145 meter.

b) Diver's horizontal motion:

Initial velocity = 2.5 m/s, acceleration = 0 $m/s^2$ , we need to calculate displacement when time is 3 seconds.

$s= 2.5*3+\frac{1}{2}*0*3^2\\ \\ =7.5meter$

So 7.5 meter far from its base the diver hit the water.

5 0

3 years ago