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3 years ago

PLEASE HELP ME SOLVE THIS!! PLZ

Physics

1 answer:

MAVERICK [17]3 years ago

5 0

Answer:

1.) Time = 3.5 seconds

2.) Range = 21 m

3.) Velocity = 34.8 m/s

Explanation:

Given that the

Height h = 60 m

Initial velocity U = 6 m/s

1.) The time taken to hit the river for a single droplet of water can be achieved by using second equation of motion

h = Ut + 1/2gt^2

Let assume that the water drops from rest. Therefore, U = 0

60 = 1/2 × 9.8t^2

60 = 4.9t^2

t^2 = 60/4.9

t^2 = 12.24

t = sqrt (12.24)

t = 3.5s

2.) The range of the projectile for a single droplet of water will be calculated by using the formula

R = Ut

Range R = 6 × 3.5 = 20.99

Range = 21 m

3.) The velocity (diagonal) when it hits the water.

Using third equation of motion

V^2 = U^2 + 2gH

Substitute values into the equation

V^2 = 6^2 + 2 × 9.8 × 60

V^2 = 36 + 1176

V^2 = 1212

V = sqrt (1212)

V = 34.8 m/s

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3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

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Answer:

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Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

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Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

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Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

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Answer:

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