Answer:
is it 3?
Explanation:
Im taking a guess and just dividing 6 and 2
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
5,000J
Explanation:
Work = Force x Distance
Distance back and forth is canceled out, so either the answer is + or -
5.0m + 5.0m = 10.0m
500N x 10.0m = 5,000J
The body system on the chart
Answer:
Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).: