h = 0.77 meter high
vertical problem:
v = g t
h = (1/2) g t^2
0.77 * 2 = 9.81 t^2
t = 0.396 second
v = 9.81 * 0.396 = 3.89 m/s = speed downward for both at ground level and the total speed for the dropped one
horizontal problem
u= 12.1 the whole time
so
for the projected one speed = sqrt(3.89^2+12.1^2) = sqrt(15.1+146.4) = 12.7
12.7- 3.9 = 8.8 m/s
The projectile launcher shown below will give the object on the right an inital horizontal speed
of 12.10 m/s. While the other object will be dropped with no initial speed. The objects are
initially 77 cm above the ground. What will be the difference in the speeds of the two objects
when they reach the ground?
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1 answer
1 answer