+0

plz plz plz help me :(

-1
73
7

2-

1. How the initial speed affecting the range and the maximum height of the projectile.

1. Which one or more of the following statements are correct?

a) The acceleration of the projectile remains constant during its upward flight.

b) The acceleration of the projectile remains constant during its downward flight.

c) The acceleration of the projectile increases during its downward flight.

d) The acceleration of the projectile decreases during its upward flight.

Oct 7, 2020

#1
+111124
+1

AND

Why are you sending us to the question with the answer already given?

Oct 7, 2020
#2
+111124
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I think of acceleration as the 'pull' on the object.

If a ball is thrown, it has an initial direction and speed.  But once it is in the air.

What direction is it being pulled?  What is pulling it?

I mean what is pulling on it to make it change its speed or direction?

Oct 7, 2020
#3
0

oh im sorry !

Oct 7, 2020
#6
+111124
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That is ok but I have asked you some questions, that you can learn from, and I am still waiting for answers.

Melody  Oct 7, 2020
#4
+10324
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1. A projectile is fired with initial speed 100 m/s at an angle 6° to the horizontal. Find the minimum  altitude.

Hello Guest!

The minimum height is the height at takeoff and landing of the projectile, so 0.

1. Which one or more of the following statements are correct?

a) The acceleration of the projectile remains constant during its upward flight.

b) The acceleration of the projectile remains constant during its downward flight.

c) The acceleration of the projectile increases during its downward flight.

d) The acceleration of the projectile decreases during its upward flight.

a) yes

b) yes

c) no

d) no

2.

How the initial speed affecting the range and the maximum height of the projectile.

$$1.\ Maximum\ height\ h_{max}$$

$$h=v\cdot t\cdot sin\ 6°-\frac{1}{2}\ gt^2$$

$$\frac{dh}{dt}=v\cdot \ sin\ 6°-g\cdot t=0$$

$$t=\frac{v\cdot sin\ 6°}{g}=\frac{100\cdot m\cdot sin\ 6°\cdot s^2}{s\cdot 9.81\cdot m}$$

$$t_{max}=1.0655\ s$$

$$h=v\cdot t\cdot sin\ 6°-\frac{1}{2}\ gt^2$$

$$h=(100m/s)\cdot (1.0655\ s)\cdot sin\ 6°-\frac{1}{2}\ (9,81m/s^2)(1.0655\ s)^2$$

$$h=11.138\ m-5.569\ m$$

$$h_{max}=5.569\ m$$

The maximum height of the projectile is 5.569 m.

$$2.\ Range\ R$$

$$t_R=2\cdot t_{max}=2\cdot 1.0655\ s$$

$$t_R=2.131\ s$$

$$R=v\cdot t_R\cdot cos\ 6°=100\frac{m}{s}\cdot 2.131\ s\cdot cos\ 6°$$

$$R=212\ m$$

The range of the projectile is 212 m.

!

Oct 7, 2020
edited by asinus  Oct 7, 2020
edited by asinus  Oct 8, 2020
edited by asinus  Oct 8, 2020
edited by asinus  Oct 8, 2020
edited by asinus  Oct 9, 2020
#7
+1

thhhaaankkss

Guest Oct 8, 2020