Notice that initial speed in the vertical is given by (25 m/s)sin(53 degree) = 2

Question

Notice that initial speed in the vertical is given by (25 m/s)sin(53 degree) = 20 m/s. The bal launched at this angle reached, the same height as the vertically launched because they have the same initial the vertical direction. In the previous part, you found that a ball fled with an initial speed of 25 m/s and an angle of 53 degree reaches the same height as a ball fled vertically with an initial speed of 20 m/s. Which ball taken longer to land? The ball fled at an angle of 53 degree stays in the air longer. Both balls are in the air the same amount of time. The ball fled vertically stays in the air longer.

Explanation / Answer

initial speed, u =25 m/s
angle of elevation, theta = 53 degree
initial speed of vertically fired ball = v = 20 m/s
same vertical height
now, for a ball dropped at height h, time taken to drop is t
h = 0.5gt^2
so, t depends only on H, for same H, same t

hence
Both the balls in the air for the same amount of time

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