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3 years ago

(50 POINT REWARD) PHYSICS, IMAGE IS ATTACHED BELOW (TRIG/VECTORS)

Physics

1 answer:

daser333 [38]3 years ago

6 0

Answer:

Explanation:

The east vector is 10cos45 = 7.07 km

The north vector is 10sin45 - 10 = -2.93 km

The displacement magnitude is

d = √(7.07² + 2.93²) = 7.65... = 8 km

θ = arctan(-2.93/7.07) = -22.51... = 23° south of east.

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3 years ago

A small, spring-loaded cannon launches a tennis ball from level ground with an initial speed vi at an angle θi with the horizont

lora16 [44]

Answer:

a. $T\ =\ \dfrac{2v_isin\theta_i}{g}$

b. $v_icos\theta_i$

c. $v_1sin\theta_i$

Explanation:

Given,

*initial velocity of the ball = $v_i$

*angle of projection = \theta_i

Horizontal component of the initial velocity of the ball = $v_x\ =\ v_icos\theta_i$

vertical initial component of the initial velocity of the ball = $v_y\ =\ v_isin\theta_i$

part a

From the kinematics,

In the y-direction motion,

total vertical displacement of the ball during the whole motion is zero.

Ball is moving under the gravitational acceleration, therefore the acceleration of the ball = -g, because gravitational acceleration always acts in the downward direction,

Let t be the time of flight of the whole motion,

$y\ =\ v_yt\ -\ \dfrac{1}{2}gt^2\\\Rightarrow 0\ =\ v_isin\theta_i t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow t\ =\ \dfrac{2v_isin\theta_i}{g}\\$

part b.

At the peak of the path of the ball, the vertical component of the velocity of the ball becomes zero, only horizontal component of the velocity acts on the ball is equal to = $v_x\ =\ v_icos\theta_i$

part c.

Initial vertical component of the velocity of the ball = $v_y\ =\ v_isin\theta_i$

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3 years ago