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BigorU [14]
3 years ago
14

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone s

ubsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:
Physics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)

Explanation:

Time=Distance/Speed

=14.5/6.63

=2.19s(to 3s.f.)

Acceleration=Final Velocity(v)-Initial Velocity(u)/Time

9.81=v-6.63/2.19

v-6.63=21.5

v=28.1m/s

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Romashka-Z-Leto [24]

The y-component of the stone's velocity when it is 8 m below the hand is 14.86 m / s

v² = u² + 2 a s

s = Displacement

u = Initial velocity

a = Acceleration

u = 8 m / s

s = 8 m

v² = 8² + 2 * 9.8 * 8

v² = 64 + 156.8

v = √ 220.8

v = 14.86 m / s

The equation used to solve the problem is an equation of motion. These equations are designed to locate an object in motion using components such as velocity, displacement, acceleration and time.

Therefore, the y-component of the stone's velocity is 14.86 m / s

To know more about Equations of motion

brainly.com/question/5955789

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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
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To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

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X_f = X_i +\frac{1}{2}(V_i+V_f)t

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In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

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PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

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Where,

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Acceleration can also be written as,

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