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3 years ago

11

Lila is a track and field athlete.She has to complete four laps around the track, which is 400 meters. The race took her 6 minut

es to complete which best describes her speed and velocity

1 answer:

hoa [83]3 years ago

3 0

V=d/t
V=?
d=400m(4)
=1600m
t=6 min.
=360 s

V=1600m/360s
V=4.4m/s

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Answer:

7.35 m/s

Explanation:

Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.

y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.

So, substituting the values of the variables into the equation, we have

y - y' = ut - 1/2gt²

0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²

- 1.2 m = 0 - (4.9 m/s²)t²

- 1.2 m = - (4.9 m/s²)t²

t² = - 1.2 m/- (4.9 m/s²)

t² = 0.245 s²

t = √(0.245 s²)

t = 0.49 s

Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.

So, d = vt

v = d/t

= 3.6 m/0.49 s

= 7.35 m/s

Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.

It is shown thus V = √(u² + v²)

= √(0² + v²)

= √(0 + v²)

= √v²

= v

= 7.35 m/s

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3 years ago

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Answer:

The statement which explains how the total time spent in the air is affected as the projectile's angle of launch increases from 25 degrees to 50 degrees is;

C. Increasing the angle from 25° to 50° will increase the total time spent in the air

Explanation:

The equation that can be used to find the total time, T, spent in the air of a projectile is given as follows;

$T = \dfrac{2 \cdot u \cdot sin(\theta) }{g}$

Where;

T = The time of flight of the projectile = The time spent in the air

u = The initial velocity of the projectile

θ = The angle of launch of the projectile

g = The acceleration due to gravity ≈ 9.81 m/s²

Given that sin(50°) > sin(25°), when the angle of launch, θ, is increased from 25 degrees to 50 degrees, we have;

Let T₁ represent the time spent in the air when the angle of launch is 25°, and let T₂ represent the time spent in the air when the angle of launch is 50°, we have;

$T_1 = \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

$T_2 = \dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g}$

sin(50°) > sin(25°), therefore, we have;

$\dfrac{2 \cdot u \cdot sin(50^{\circ}) }{g} > \dfrac{2 \cdot u \cdot sin(25^{\circ}) }{g}$

Therefore;

T₂ > T₁

Therefore, increasing the angle at which the projectile is launched from 25° to 50° will increase the total time spent in the air.

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2 years ago