Physics, 26.09.2019 13:20 zionboy77

(b) a ball is thrown from a point 1.50 m above the ground. the initial velocity is 19.5 m/s at

an angle of 30.0˚ above the horizontal.

(i) find the maximum height of the ball above the ground. [6.35 m] [2 marks]

(ii) what is the speed of the ball just before it strikes the ground? [20.2 m/s] [3 marks]

(b) a ball is thrown from a point 1.50 m above the ground. the initial velocity is 19.5 m/s at

Answers: 1

Show answers

Answers

Answer from: kee387

The answers are:

(i) 6.35m

(ii) 20.2 m/s

It seems like you already have the answer, but let me show you how to get it:

You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)

When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components.

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = 9.75 m/s

Vix = CosΘVi = (Cos30°)(19.5m/s) = 16.89 m/s

Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.

(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

Where:

Vf = final velocity

Vi = inital velocity

a = 9.8m/s²

d = displacement

We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

$dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}$

dy = 4.85m

So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT ABOVE THE GROUND.

4.85m + 1.50m = 6.35m

(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$

$Vfy^{2} = 0m/s^{2}+2(9.8m/s^{2}(6.35m)$

$\sqrt{Vfy^{2}} = \sqrt{124.46m^{2}/s^{2}}$

Vfy= 11.16m/s

OKAY! We are at our last step. Now to get the resultant velocity, we apply the Pythagorean theorem.

$Vf^{2} = Vfx^{2} + Vfy^{2}$

$\sqrt{Vf^{2}} = \sqrt{(16.89m/s)^{2}+(11.16m/s)^{2}}$

Vf =20.2m/s

The ball was falling at 20.2m/s before it hit the ground.

Answer from: Quest

An atom (ours, not carroll's) consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. the positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. although the nucleus is heavy, it is quite small compared with the overall size of an atom. the radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6 nm. * if an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.read more on -

Answer from: Quest

volts times amps 6 x 0.5 3w

Another question on Physics

Physics, 22.06.2019 10:00

The rocket is fired vertically and tracked by the radar station shown. when θ reaches 66°, other corresponding measurements give the values r = 32700 ft, r¨ = 85 ft/sec2, and θ˙ = 0.019 rad/sec. calculate the magnitudes of the velocity and acceleration of the rocket at this position.

Answers: 3

Answer

Physics, 22.06.2019 11:00

Need this now! do pilots live long

Answers: 1

Answer

Physics, 22.06.2019 12:30

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. when you push the platform, you compress the springs. you do an amount of work of 79.0 j when you compress the springs a distance of 0.230 m from their uncompressed length. (a) what magnitude of force must you apply to hold the platform in this position? (b)how much additional work must you do to move the platform a distance 0.230 m farther? (c) what maximum force must you apply to move the platform a distance 0.230 m farther?

Answers: 1

Answer

Physics, 22.06.2019 16:00

Two balls, each with a mass of 0.5 kg, collide on a pool table. is the law of conservation of momentum satisfied in the collision? explain why or why not

Answers: 1

Answer

You know the right answer?

(b) a ball is thrown from a point 1.50 m above the ground. the initial velocity is 19.5 m/s at

Questions