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Gnesinka [82]
3 years ago
5

A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?

Physics
2 answers:
Kazeer [188]3 years ago
4 0
-- She went up for 0.4 sec and down for 0.4 sec.

-- The vertical distance traveled in gravity during ' t ' seconds is

                   D  =  (1/2)  x  (g)  x  (t)²

                       = (1/2) (9.8 m/s²) (0.4 sec)²

                       =    (4.9 m/s²)  x  (0.16 s²)

                       =      0.784 meter        ( B )
Alinara [238K]3 years ago
4 0
After 0.40 seconds, she must have reached her max height
she spends half the time going up and the other half falling down

t = 0.40
vf = 0
a = -9.8

y = vf t - 1/2 a t^2

y = 0 - 1/2 (-9.8) (0.4)^2

y = 4.9 (0.4)^2

Y=0.784

This will be your answer.
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3 years ago
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Carlos is making phosphorous trichloride using the equation below. He adds 15 g of phosphorus.
Degger [83]
Given:

The balanced chemical reaction of the synthesis of phosphorus trichloride:

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An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fa
sineoko [7]

Answer:

v = 83.1 % of speed of light

Explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

T_s = T_e \times \gamma

where c is the speed of light

v is the speed of the rock star moving

T_s = T_e\times \sqrt{1-\dfrac{v^2}{c^2}}

1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}

\sqrt{1-\dfrac{v^2}{c^2}} =0.556

squaring both side

1-\dfrac{v^2}{c^2}=0.3086

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7 0
3 years ago
A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif
Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

R=\frac{v^{2}sin(2\theta)}{g}

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

gR=v^{2}sin(2\theta)

\frac{gR}{v^{2}}=sin(2\theta)

sin^{-1}(\frac{gR}{v^{2}})=2\theta

\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}

so now we can substitute the given data:

\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

tan \theta = \frac{h}{45.5m}

so we can solve this for h, so we get:

h=45.5m*tan(0.05614^{o})

which yields:

h=0.0445m

or

h=4.45cm

5 0
3 years ago
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