All categories

1 year ago

This is being graded.Great answers only

Physics

1 answer:

navik [9.2K]1 year ago

6 0

Answer:

Explanation:

When a ball is at its highest point, its acceleration is 0. The only thing pulling it down, is gravity so, the correct option is D. Downwards.

I'm not sure however due to the fact that the arrows on the other options are facing backwards meaning the ball would go backwards; and for the one option the ball is going up. However, the ball isn't going backwards nor up, so it must be D. If I am wrong I greatly apologize.

You might be interested in

10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball

SVEN [57.7K]

Answer:

Josie's ball faster than T

3 0

1 year ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

3 years ago

How is velocity and instantaneous speed alike

allsm [11]

<span>velocity is defined as the rate of change of displacement irrespective of the length of the path travelled while speed is the average rate of covering distance. but in the liming case where the instantaneous velocity is given as v=dx/dt where dx is the small displacement in a small interval dt, both the speed and velocity have the same magnitude and the direction of velocity is the direction of the tangent to the corresponding displacement-time curve.</span>

5 0

3 years ago

In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th

Arlecino [84]

We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon

8 0

2 years ago

As speed (velocity) increases, potential energy increases true or false

Lyrx [107]

Answer:

true

Explanation:

3 0

2 years ago

This is being graded.Great answers only​

This is being graded.Great answers only