Positive 2n to the right of the box
<span>2.56 sec (t=1/0.39)
f=3.9/10</span>
Vx=cos60(4)
x-component of velocity
<span>If you think about it, it makes a right triangle when you combine all the different types of forces together such as v, vx and vy. Then, you can use trigonometry and soh cah toa in order to figure out vx. </span>
Answer:
2.65m/s
Explanation:
Using the equation of motion:
v² = u²+2a∆S where
v is the final velocity
u is the initial velocity
∆S is the change in distance
a is the acceleration
Given
u = 0m/s
a = 9.8m/s²
∆S = 1.3-0.943
∆S = 0.357m
Substituting the given parameters into the formula
v² = 0²+2(9.8)(0.357)
v² = 0+6.9972
v² = 6.9972
v=√6.9972
v = 2.65m/s
Hence the velocity at which it hit the ground is 2.65m/s
B) Applied and gravitational forces
