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

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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

nah. i'm just sleepy.

Answer from: Quest

compass can be used day and night, cloudy or sunny

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

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