Asked by Jess
If a projectile is fired straight up at a speed of 10 m/s, the total time to return to its starting position is about ?
-Hfinal=hinitial+ vinitial*t -4.9t^2
0=0+10t-4.9t^2 when I solved for t I got .49t. Also, where did 4.9 come from?
A man leans over the edge of a cliff and throws a rock upward at 4.9m/s. How far below the level from which it was thrown is the rock 2 seconds later?
-Hfinal=Hinitial+Vi*t - 4.9t2
hfinal=0+4.9m/s*2 s -4.9t^2, then I get hfianl=9.8m/s/s-4.9^2, I don't dont what to do next?
-Hfinal=hinitial+ vinitial*t -4.9t^2
0=0+10t-4.9t^2 when I solved for t I got .49t. Also, where did 4.9 come from?
A man leans over the edge of a cliff and throws a rock upward at 4.9m/s. How far below the level from which it was thrown is the rock 2 seconds later?
-Hfinal=Hinitial+Vi*t - 4.9t2
hfinal=0+4.9m/s*2 s -4.9t^2, then I get hfianl=9.8m/s/s-4.9^2, I don't dont what to do next?
Answers
Answered by
drwls
4.9 is g/2, and g is the acceleration of gravity on Earth. The answer should be in seconds, not 0.49 t. You also did the calculation incorrectly.
In your second problem, the equation to solve is
y = 4.9t - 4.9 t^2. Plug in t = 2s and solve for y.
y(at t=2s) = 9.8 - 4*9.8 = -29.4 m
In your second problem, the equation to solve is
y = 4.9t - 4.9 t^2. Plug in t = 2s and solve for y.
y(at t=2s) = 9.8 - 4*9.8 = -29.4 m
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