A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be v=7.6i^+6.1j^ in metre per second ( i^ is horizontal, j^ is upward). Give the approximate answers. What is the magnitude of the initial velocity of the ball?

ux=vx=7.6 m/s vy=uy2-2gh uy=vy2+2gh=(6.1)2+2×10×9.1=14.8 m/s u=ux2+uy2=(7.6)2+(14.8)2=16.6 m/s

A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be v=7.6i^+6.1j^ in metre per second ( i^ is horizontal, j^ is upward). Give the approximate answers. What is the magnitude of the initial velocity of the ball?

ux=vx=7.6 m/s vy=uy2-2gh uy=vy2+2gh=(6.1)2+2×10×9.1=14.8 m/s u=ux2+uy2=(7.6)2+(14.8)2=16.6 m/s

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A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be $v = 7.6 \hat{i} + 6.1 \hat{j}$ in metre per second ( $\hat{i}$ is horizontal, $\hat{j}$ is upward). Give the approximate answers. What is the magnitude of the initial velocity of the ball?

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18.6 m/s

15.6 m/s

16.6 m/s

17.6 m/s

answer is B.

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Detailed Solution

$u_{x} = v_{x} = 7.6 m / s v_{y} = \sqrt{u_{y}^{2} - 2 g h} u_{y} = \sqrt{v_{y}^{2} + 2 g h} = \sqrt{(6.1)^{2} + 2 \times 10 \times 9.1} = 14.8 m / s u = \sqrt{u_{x}^{2} + u_{y}^{2}} = \sqrt{(7.6)^{2} + (14.8)^{2}} = 16.6 m / s$

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A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be v=7.6i^+6.1j^ in metre per second ( i^ is horizontal, j^ is upward). Give the approximate answers. What is the magnitude of the initial velocity of the ball?

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A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be $v = 7.6 \hat{i} + 6.1 \hat{j}$ in metre per second ( $\hat{i}$ is horizontal, $\hat{j}$ is upward). Give the approximate answers. What is the magnitude of the initial velocity of the ball?