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AURORKA [14]
2 years ago
7

n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It

travels a horizontal distance of 40.0 mm. Neglect air resistance. Part A How long is the ball in the air
Physics
1 answer:
Pani-rosa [81]2 years ago
4 0

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

v_x = RT\\\\T = \frac{v_x}{R}

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

v_x = horizontal speed = 9 m/s

Therefore,

T = \frac{9\ m/s}{40\ m}

<u>T = 0.225 s</u>

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A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

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7 0
1 year ago
Match the following ERA’s Question 4
inna [77]
1. Cenozoic ERA
2. Mesozoic ERA
3. Paleozoic ERA
8 0
2 years ago
A ray of light strikes a plane mirror at an angle of incidence 35°. If the mirror is rotated through 10°.
xxTIMURxx [149]

Answer:

1 35°

2 20°

3 50°

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7 0
3 years ago
A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc
BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

t=\frac{usin\alpha }{g}

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

t=40sin30/9.81\\t=2.0secs

b. the expression for the maximum height is expressed as

H=\frac{u^{2}sin^{2}\alpha  }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0
2 years ago
Please help me if you know physics
Nitella [24]

Answer:

  19.  down

  20.  True

Explanation:

The examples I saw were projectiles launched at different angles and speeds. In each case, the only acceleration was ...

  19:  down

  20:  due to gravity (true)

4 0
2 years ago
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