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Elina [12.6K]
3 years ago
7

a player passes a basketball to another player who catches it at the same level from which it was thrown. the initial speed of t

he ball is 7.1m/s and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) it's time of flight?
Physics
2 answers:
Talja [164]3 years ago
5 0

Answer: horizontal direction ; Time of flight = 0.65seconds.

Explanation:

Initial direction: horizontal direction.

Time of the flight= velocity is (distance/time)

Speed= 7.1m/s

Distance= 4.6m

Time= X

Time = (Distance/speed)

X = (4.6 / 7.1) = 0.65 seconds

Time of flight = 0.65seconds.

Olegator [25]3 years ago
4 0
A is this answer bc i dk so yea

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a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin
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Answer:

a) E_{p} = 0

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E_{m} = 168.7 J

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c) E_{p} = 169.2 J

E_{k} = 0

E_{m} = 169.2 J

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

E_{p} = mgh = 0

The kinetic energy is:

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

E_{p} = mgh = 1.5*9.81*5 = 73.6 J

Now, to find the kinetic energy we need to calculate the speed at 5 m:

v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9

v_{f} = \sqrt{126.9} = 11.3 m/s

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J

c) At its maximum height:

v_{f}: is the final speed = 0

h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m

Now, the potential, kinetic and mechanical energies are:

E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J

E_{k} = \frac{1}{2}mv^{2} = 0

E_{m} = 169.2 J + 0 = 169.2 J

I hope it helps you!    

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