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

Question

A ball is tossed from an upper-story window of a building. Theball is given an initial velocity of 8 m/2 at an angle of 20degreesbelow the horzontal. It strikes the ground 3.00s later. a) How far horizaontally from the base of the building doesthe ball strike the ground? b) Find the height from which the ball was thrown. c) How long does it take the ball to reach a point 10m belowthe level of launching? A ball is tossed from an upper-story window of a building. Theball is given an initial velocity of 8 m/2 at an angle of 20degreesbelow the horzontal. It strikes the ground 3.00s later. a) How far horizaontally from the base of the building doesthe ball strike the ground? b) Find the height from which the ball was thrown. c) How long does it take the ball to reach a point 10m belowthe level of launching?

Explanation / Answer

X = Vx(T) where X = horizontal distance from base of building where ball willland Vx = horizontal component of velocity T = time ball strikes the ground after being tossed Substituting appropriate values, X = 8(cos 20)(3) X = 22.55 m Y = Vy(T) + (1/2)(g)T^2 Y = height from which ball was thrown Vy = vertical component of the velocity T = time when ball strikes the ground g = acceleration due to gravity = 9.8 m/sec^2 (constant) Substituting appropriate values, Y = 8(sin 20)(3) + (1/2)(9.8)(3^2) Y = 52.31 meters Formula is S = Vy(T) + (1/2)(g)T^2 where S = 10 m Vy = 8(sin 20) g = 9.8 m/sec^2 T = time for ball to reach a point 10 m below the launchinglevel Substituting values, 10 = 8(sin 20)T + (1/2)(9.8)T^2 Rearranging the above, 1/2)(9.8)T^2 + 8(sin 20)T - 10 = 0 Solving for "T" using the quadratic formula, T = 0.22 sec.

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