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3 years ago

A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be

Physics

1 answer:

Mariulka [41]3 years ago

5 0

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t :

Equations of the uniformly accelerated rectilinear motion of upward (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m

y: vertical position in meters (m)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the initial vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 = (v₀y)²

$v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }$

v₀y = 14.3 m/s

Calculation of the maximum height the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball (R)

We replace data in the equation (1)

x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

$v= \sqrt{v_{x}^{2}+v_{y}^{2} }$

$v= \sqrt{(7.7)^{2}+ (-14.3)^{2} }$

v= 16,2 m/s

$\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })$

$\alpha = tan^{-1} (\frac{-14.3 }{7.7 })$

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

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Answer:

Part A:

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Part B:

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$\lambda = \frac{h}{mv}$ (2)

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But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

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Case for the proton:

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$\lambda = 6.05x10^{-14}m$

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