A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° abo

Question

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground and experience negligible air resistance.

(a) What is the magnitude of the ball's velocity just before it hits the ground?

(b) At what angle below the horizontal does the ball approach the ground?

one thing I was confused about.. while solving we chose that vx= vox which is 8.. should't it be ( vx =8cos40) since vox is the component? hope you got my point.....

Explanation / Answer

(A) initial vertical velocity, v0y = 8 sin40 = 5.14 m/s

ay = - 9.81 m/s^2

y = yf - yi = 0 - 1 = - 1m

Applying vf^2 - v0^2 = 2 a d

vfy^2 - 5.14^2 = 2(-9.81)(-1)

vfy = -6.78 m/s

inhorizontal, a_x = 0

hence vfx = v0x = 8 cos40 = 6.13 m/s

vf = 6.13i - 6.78j m/s

(a) magnitude = sqrt(vfx^2 + vfy^2) = 9.14 m/s

(b) theta = tan^-1(6.78 / 6.13) = 48 deg

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