A football is punted at 22.0 m/s at an angle of 31.0degree above the horizon. As

Question

A football is punted at 22.0 m/s at an angle of 31.0degree above the horizon. Assume it starts 1.00 m above ground level. (Neglect any effects due to air resistance.) What is the magnitude of v_x of the ball when it is 5.00 m above ground level? (Express your answer to three significant figures.) You currently have 2 submissions for this question. Only 4 submission are allowed. You can make 2 more submissions for this question. What is the magnitude of v_y of the ball when it is 5.00 m above ground level? (Express your answer to three significant figures.) you currently have 2 submissions for this question. Only 4 submission are allowed. you can make 2 more submissions for this question.

Explanation / Answer

here

the initial speed of ball , u = 22 m/s

the angle theta = 31 degree

ux = u * cos(theta)

ux = 22 * cos(31)

ux = 18.86 m/s

uy = u * sin(theta)

uy = 22 * sin(31)

uy = 11.3 m/s

height of ball , h = (5-1) = 4 m

let the vertical component of velocity at h = 4 m be vy

using first equation of motion

vy^2 - uy^2 = - 2 * g * h

vy^2 - 11.33^2 = - 2 * 9.8 * 4

vy = 7.07 m/s

and

vx = 18.86 m/s

the velocity vector of ball when it is 5 m above the ground is (18.86 i + 7.07 j) m/s

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