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3 years ago

Will upvote!!!!

Physics

2 answers:

Aloiza [94]3 years ago

7 0

Potential Energy is correct. I took the test

juin [17]3 years ago

4 0

D. potential energy, because there is a bunch of water pent up, essentially stationary, waiting to roll down the steep mountain from the peak, so to say. if the dam were to be removed it would become kinetic.

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You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of th

solmaris [256]

a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

$v_x = u cos \theta$

where

u = 8.3 m/s is the initial velocity of the balloon

$\theta=27^{\circ}$ is the angle of projection

Substituting,

$v_x = (8.3)(cos 27^{\circ})=7.4 m/s$

The vertical component of the initial velocity of a projectile is given by

$u_y = u sin \theta$

where

u is the initial velocity

$\theta$ is the angle of projection

Here we have

u = 8.3 m/s

$\theta=27^{\circ}$

Substituting,

$u_y = (8.3)(sin 27^{\circ})=3.8 m/s$

The vertical component of the velocity of the balloon follows the suvat equation

$v_y = u_y - gt$

where

$v_y$ is the vertical velocity at time t

$u_y = 3.8 m/s$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

$v_y = 0$

So we get:

$0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s$

The maximum height of the balloon can be calculated using the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where

$u_y = 3.8 m/s$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

$s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m$

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

$t=\frac{2u_y}{g}$

where

$u_y$ is the initial vertical velocity

$g$ is the acceleration of gravity

Here we have

$u_y = 3.8 m/s$

$g=9.8 m/s^2$

Substituting,

$t=\frac{2(3.8)}{9.8}=0.78 s$

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

$d=v_x t$