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xxTIMURxx [149]
3 years ago
11

Your throw a ball straight upward at an initial speed of 5 m/s. How many times does the ball pass a point 2 m above the point yo

u launched it from? Draw an x-t and a v-t diagram for the motion of this ball.

Physics
1 answer:
Lina20 [59]3 years ago
8 0

Answer:

The ball never passes 2m high, Hmax=1.27m

Explanation:

we assume the ball doesn't bounce when it hits the ground.

We calculate the maximum height, Vf = 0.

v_{o}^{2}=2gH_{max}\\H_{max}=v_{o}^{2}/(2g)=5^{2}/(2*9.81)=1.27m

So, the ball never passes 2m high.

Kinematics equations:

x(t)=v_{o}t-1/2*g*t^{2}\\v(t)=v_{o}-gt

Find annexed the graphics of x(t) and v(t)

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The solution for this problem would be:(10 - 500x) / (5 - x) 
so start by doing: 
x(5*50*2) - xV + 5V = 0.02(5*50*2) 
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V(5 - x) = 10 - 500x 
V = (10 - 500x) / (5 - x) 
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1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

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The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

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