As per given conditions there are two directions along which forces are acting
1. Net force along left direction is given as

2. Net force towards right direction is given as

now since the two forces here in opposite direction so here we will have net force given as



so here net forces must be 440 N towards right
Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.
<em>projectile can only follow the straight line path when it is launched upward straightly so the correct option is <u>90 degree with respect to horizontal x -axis ..:)</u></em>
The answer for this change in the magnitude of momentum is the same for both because momentum is always conserved so both vehicles have the identical change.
So for determining who has the greater change in kinetic energy, momentum (P) = mv so P^2 = m^2 v^2 P^2 / 2m = 1/2 m v^2 = energy So the weightier the mass the smaller the energy change for the same momentum change so in here, the car has a greater change in kinetic energy.
Explanation:
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