A ball is tossed from an upper-story window of a building. The ballis given an i

Question

A ball is tossed from an upper-story window of a building. The ballis given an initial velocity of 8.80m/s at an angle of 25.0° below the horizontal. It strikes the ground6.00 s later. (a) How far horizontally from the base of thebuilding does the ball strike the ground?
1 m

(b) Find the height from which the ball was thrown.
2 m

(c) How long does it take the ball to reach a point 10.0 m belowthe level of launching?
3 s
(a) How far horizontally from the base of thebuilding does the ball strike the ground?
1 m

(b) Find the height from which the ball was thrown.
2 m

(c) How long does it take the ball to reach a point 10.0 m belowthe level of launching?
3 s

Explanation / Answer

x = v0*t = (8.80*cos(25))*6s x = 47.9 m
b) v_y(t) =v_y(0) + a*t = 8.80*sin(25 degrees) + 9.8 m/s * 6s =81.11 m/s
v_y^2 = v_y(0)^2 + 2*y*a
y = (v_y^2 - v_y(0)^2)/2a = (81.11^2 -[8.80sin(25)]^2)/(2*9.8) y = 198.7 EDIT: Poster below has it correct
c)
v_y^2 = v_y(0) + 2*y*a v_y^2 = (8.80*sin(25))^2 + 2*10*9.8 v_y^2 = 209.831 v_y = 14.49
v_y = v_0 + a*t
t = (v_y-v_0)/a = (14.49 - 8.8*sin(25))/9.8 t = 1.10

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