a = (delta v)/t = (v - vo)/t = (25 m/s - 15 m/s)/(4 s) = 2.5 m/s2 So, your average velocity = distance/time = (2 km)/(1/20 h + 1/30 h) = (2 km)/(1/12 h) = 24 km/h
d = 0.75 m, a = 10 m/s2, t = ? The hang time is twice the time to fall = 2(0.385 s) = 0.77 s First, you need to realize that the salmon goes up some distance d (that we want to know) and then falls from rest that same distance, so we can use the equation v = 5 m/s, a = 10 m/s2, t = ? Since v = at, t = v/a = (5 m/s)/(10 m/s2) = 0.5 s Step 2: Find the distance fallen from rest in 0.5 s a = 10 m/s2, t = 0.5s, d = ?
If it is thrown at an angle, at the top of its path, its vertical velocity will be zero, however its horizontal velocity will be the same as its initial horizontal velocity minus whatever loss in speed as a result of air friction at that point. We won't know what that is without more information. |