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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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6 0
3 years ago
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
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Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

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hammer [34]

Answer:

a_M=1.92a_A

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\omega_M=\omega_A = Angular speed

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\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A

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