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:
A = 1.4 m/s²
B = -0.10493 m/s³
a = 1.29507 m/s²
T = 28095.8271 N
T = 1.13198 W
Explanation:
t = Time taken
g = Acceleration due to gravity = 9.81 m/s²
The equation
Differentiating with respect to time
At t = 0
Hence, A = 1.4 m/s²
B = -0.10493 m/s³
At t = 5 seconds
a = 1.29507 m/s²
T = 28095.8271 N
Weight of rocket
T = 1.13198 W
Answer:
It will double.
Explanation:
Newton's Law of Gravity states that 
, where G is the gravitational constant, <em>r</em> is the distance between the objects' centers, and 
and 
are the objects' masses. We just have simple math here: by doubling , we double the entire fraction. By doubling the entire fraction, we double the gravitational pull. Therefore, <em>Newton's Law of Gravity states that if the mass of one object doubles, the gravitational pull on a second object will </em><em>double</em><em>.</em>
I hope this helps you understand it! Have a great day, 'kay?
Explanation:
Below is an attachment containing the solution.
We can use the formula Vf^2= Vi^2 + 2ad.
So,
Vi = 0 since we are dropping the object
Therefore,
Vf^2 = 2 (9.8) (49)
You can do the calculator work.
