How high can a body vertically thrown with a speed of 40m/s raise after 3 sec (neglecting air

Physics

1 answer:

Tcecarenko [31]3 years ago

5 0

y = 75.9 m

Explanation:

y = -(1/2)gt^2 + v0yt + y0

If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.

y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)

= -44.1 m + 120 m

= 75.9

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