Q5) At a diving competition, the diving board is 3.0 m above the surface of the

Question

Q5)   At a diving competition, the diving board is 3.0 m above the surface of the pool. The diver gets a running start and jumps into the air at an angle at the end of the diving board to begin his dive. The vertical component of the velocity of his jump is 1.9 m/s. When he lands in the water, he has traveled 1.5 m in the horizontal direction.

A) Find the amount of time that the diver was in the air.?

B) Find the magnitude and direction of his initial velocity.?

C) Find the speed and direction he is traveling when he lands in the pool.?

Explanation / Answer

here,

height of the board , h = 3 m

a)

let the time taken to hit the water be t

- h = uy * t - 0.5 * g * t^2

- 3 = 1.9 * t - 4.9 * t^2

t = 1 s

the time taken is 1 s

b)

let the initial velocity be v and the angle of launch be theta

u * sin(theta) = uy = 1.9 ...(1)

and

u * cos(theta) * t = x

u * cos(theta) * 1 = 1.5 ....(2)

from (1) and (2)

u = 2.42 m/s and theta = 51.7 degree

b)

final vertical velocity , vy = uy - g * t

vy = - 7.9 m/s

vx = 1.5 m/s

|v| = sqrt(1.5^2 + 7.9^2) = 8.04 m/s

theta = arctan(-7.9/1.5) = 79.2 degree below the horizontal

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