The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.
<h3>
What is vertical motion of a projecile?</h3>
The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;
Vy = Vsinθ
<h3>
What is horizontal motion of a projecile?</h3>
The horizontal velocity of a projectile is given by the following formula;
Vx = Vcosθ
<h3>Direction of the motion</h3>
The direction of the motion is calculated as follows;
tanθ = Vy/Vx
Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.
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Answer:
Explanation:
The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.
We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed
(1)
We solve in order to find the receiver mass:
N2 = 3*n1
T2 = 2*T1
V1 = V2
(n2 * T2)/P2 = (n1 * T1)/P1
3 n1 * 2 T1 / P2 = n1 *T1 / P1
P2 = 6*P1
Since P2 is 6P1, it is 6 times greater than original pressure
Explanation:
If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.
Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m