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An elevator is moving upward at a constant speed of 4 m/s. A man stand- ing 10 m above the top of the elevator throws a ball upward with a speed of 3 m/s. Determine (a) when the ball will hit the elevator, (b) where the ball will hit the elevator with respect to the location of the man.

Respuesta :

Answer:

The ball will hit the elevator in 1.33 sec and the ball will hit the elevator is 4.68 m below the man.

Explanation:

Given that,

Constant speed of elevator = 4 m/s

Distance = 10 m

Speed of ball = 3 m/s

We need to calculate the time

We will calculate the vertical speed of elevator

[tex]y(e)=4t-10[/tex]....(I)

Here, speed is constant so g will neglect

We will calculate the vertical speed of ball

[tex]y(b)=-\dfrac{1}{2}gt^2+ut[/tex]

[tex]y(b)=-4.9t^2+3t[/tex]....(II)

From equation (I) and (II)

[tex]4t-10=-4.9t^2+3t[/tex]

[tex]4.9t^2+t-10=0[/tex]

[tex]t = 1.33\ sec[/tex]

Negative time will neglect

(b). We need to calculate the distance

Put the value of t in equation (I)

[tex]y(e) =4\times1.33-10[/tex]

[tex]y(e)=-4.68\ m[/tex]

Hence, The ball will hit the elevator in 1.33 sec and the ball will hit the elevator is 4.68 m below the man.

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