Question

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

resistance)​

Answer 1

y = 75.9 m

Step-by-step 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