Question

A ball is thrown from a point 1.10 m above the ground. The initial velocity is 19.7 m/s at an angleabove the horizontal. If the maximum height of the ball aboveground is 7.29 m, what was the angle atwhich the ball was thrown? Answer in degrees. [Hint:all information necessary to solve this problemis providedl.]

Answer #1

let the angle = =?

u = 19.7 m/s

H = 7.29 and h = 1.10 m

uy = usin = 19.7 sin

in y direction ( take up as +ve)

heightest point (S)= H - h = 7.29 -1.1 = 6.19 m

at this point Vy = 0

so we can write

from Vy^{2} = Uy^{2} + 2aS

0 = (19.7sin)^{2}
- 2g6.19

(19.7sin)^{2}
= 2g6.19

sin = 0.559

= 34 degree

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