Compute first for the vertical motion, the formula is:
y = gt²/2
0.810 m = (9.81 m/s²)(t)²/2
t = 0.4064 s
whereas the horizontal motion is computed by:
x = (vx)t
4.65 m = (vx)(0.4064 s)
4.65 m/ 0.4064s = (vx)
(vx) = 11.44 m / s
So look for the final vertical speed.
(vy) = gt
(vy) = (9.81 m/s²)(0.4064 s)
(vy) = 3.99 m/s
speed with which it hit the ground:
v = sqrt[(vx)² + (vy)²]
v = sqrt[(11.44 m/s)² + (3.99 m/s)²]
v = 12.12 m / s
B. is it i just got done this in class like two weeks ago hope it helps
The two letters are B and A, in that order.
Average speed = distance travelled / time used
Answer:
10.21 N
Explanation:
As the force is a vector, it can be decomposed in two components perpendicular each other, so there is no projection of one component in the direction of the other.
When divided in this way, the magnitude of the resultant vector can be found simply applying trigonometry, as follows:
F² = Fx² + Fy² ⇒ F = √(Fx)²+(Fy)²
Replacing by Fx= 5.17 N and Fy = 8.8 N, we get:
F = √(5.17)²+(8.8)² =10.21 N
