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Step2247 [10]
3 years ago
7

A ball is projected upward at time t = 0.0 s, from a point on a roof 60 m above the ground. The ball rises, then falls until it

strikes the ground. The initial velocity of the ball is At time what is the acceleration of the ball? Neglect air resistance.
Physics
1 answer:
musickatia [10]3 years ago
7 0
<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here

then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))

The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>
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Answer:

Yes

Explanation:

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What cell organelle is necessary for cellular respiration
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It is the cytoplasm.
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Read 2 more answers
8
Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, \overrightarrow{v_1}=-120\ \vec{i}

Velocity in north direction is, \overrightarrow{v_2}=120\ \vec{j}

Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
A ball is thrown vertically upwards with a velocity of 30m/s. Determine the maximum height reached
padilas [110]
The maximum height reached is 45.92 m

6 0
2 years ago
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