(8c4p37) A ball is shot from the ground into the air. At a height of 9.6 m, the

Question

(8c4p37) A ball is shot from the ground into the air. At a height of 9.6 m, the velocity is observed to be v = (7.8)i + (7.1)] in meters per second (i horizontal, j upward). To what maximum height will the ball rise? Submit Answer Tries 0/8 What will be the total horizontal distance traveled by the ball? led by the bali? Submit Answer Tries 0/8 What is the velocity of the ball (magnitude and direction) the instant before it hits the ground? That is what is its i-component? Submit Answer Tries 0/8 What is its j-component? Submit Answer Tries 0/8

Explanation / Answer

According to the concept of the vector algebra

Given that

Velocity v=7.8i+7.6j=Ux i+Uy j

Now we find the maximum height

Maximum height Hmax=(Uy ) ^2/2g=7.1^2/(2*9.8)=2.6 m

Now we find the range

Range =2Ux*Uy/g

=2*7.8*7.1/9.8

=11.3 m

Now we find the x-component velocity at position of hitting the ground

In this case acceleration a=o

Vx=(ux+2gh)^1/2=(7.8^2+0)^1/2=7.8 m/s

Velocity at y-component Vy=(uy^2+2gh)^1/2

=(7.1^2+2*9.8*9.6)^1/2

=15.4 m/s

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