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.Am I suppose to plug .49t back into the equation?
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?
-For this problem I am suppose to get 9.8m/s and instead got -29.4 m
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
-Hfinal=hinitial+ vinitial*t -4.9t^2
0=0+10t-4.9t^2 when I solved for t I got .49t.Am I suppose to plug .49t back into the equation?
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?
-For this problem I am suppose to get 9.8m/s and instead got -29.4 m
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
Answers
Answered by
bobpursley
The first is a quadratic. Either factor it, or use the quadratic equation.
On the second, you cannot get 9.8 m/s as a distance, that is a velocity.
On the second, you cannot get 9.8 m/s as a distance, that is a velocity.
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