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notka56 [123]
3 years ago
10

A ball is launched vertically upward from the edge of a cliff. The ball reaches its maximum height 1.6 seconds after launch. Bar

ely missing the edge of the cliff as it falls downward, the ball strikes the ground 6 seconds after being launched. a) What was the ball's initial velocity? b) What is the maximum height the ball reached above the cliff? c) How tall is the cliff?
Physics
1 answer:
irakobra [83]3 years ago
8 0

Answer: a) 60m/s b) 360m c) 180m

Explanation:

Projectile motion occurs when an object launched into space falls freely under the influence of gravity.

If the time of flight (T) = 2Usin(theta)/g where;

U is the velocity of the object

theta is the angle at which the object was thrown

g is the acceleration due to gravity

Given T = 6 seconds

theta= 90°(object launched upwards)

g = 10m/s²

U= ?

Substituting in the formula we have;

6 = Usin90°/10

60 = Usin90°

since sin90° = 1

U = 60m/s

The balls initial velocity is 60m/s.

b) Maximum Height reached (H) = U²sin²theta/g

Given U = 60m/s, theta = 90° g = 10m/s²

H = 60²(sin90°)²/10

H = 3600/10

H = 360m.

The maximum height reached is 360m

c) at maximum height, v = 0

Using v² = u² - 2gh

0 = 60² - 2(10)h

-3600 = -20h

h = 3600/20

h = 180m

The cliff is 180m tall

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In this case, because the Initial Velocity of the ball is a vecotr, it can be defined using its two directionals compounds. One, on the Y-axis and another on the X-axis. This can be related, on how a hypotenuse is the product of two legs of the triangle. Because of this, each one of this, can be know using the following equation:

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Where Vo is the initial velocity 27.5 m/s, and ∅ is the angle which is 20.5°. So we calculate:

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To know how tall does the baseball hit the wall, we need to know first how much time it takes the ball to reach the wall on the X-axis. The wall is 17.5m away, we velocity on Vx that is constant we can calculate as it follow:

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Know with the time, we can calculate the how tall it got on that time with the following equation:

x = (Vo*t) + (\frac{1}{2} *a*t^{2} )

Where Vo is the Y-Compound of the Initial Velocity.

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x = (Vy*T) + (\frac{1}{2} *g*T^{2} )

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