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A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building. Part A If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?

Respuesta :

Answer:

H = Vy t - 1,2 g t^2            formula for height of ball after t sec

H = 10 - 1.3 = 8.7 m

Vy = 27 sin 47 = 19.7 m/s     vertical speed of ball

8.7 = 19.7 t - 9.8/2 t^2       height of ball after t sec

4.9 t^2 - 19.7 + 8.7 = 0       rearranging

[19.7 ± (388 - 170)^1/2] / 2 *4.9 = [19.7 ± 14.7] / 9.8 = .51 ,3.5 sec

.51 sec would be on the way up and 3.5 sec on the way down

Sx = 27 * cos 47 * 3.5 = 64.4 m around 200 ft seems reasonable