The greater the force on an object the greater the distance while the less the force the less the distance.
Answer:
Explanation:
As we know that there is no external horizontal force on the cannon ball + performer
So here total momentum of the system must be conserved
so we will have
now from above equation we have
In the projectile motion, the direction is characterized by a shape of the arc. Its horizontal component travels in constant velocity while the vertical component travels in constant acceleration. The equation to be used is:
2ay = |v² - v₀²|
where
a is the acceleration due to gravity equal to 9.81 m/s²
y is the height
v is the final velocity
v₀ is the initial velocity
Substitute y=1/2*H and v = 3/4*v₀. The equation for maximum height is
H = v₀²sin²θ/2a
Thus,
(2)(9.81)(1/2)(H) = |(3/4v₀)² - v₀²|
(2)(9.81)(1/2)(v₀²sin²θ/2(9.81)) = |(3/4v₀)² - v₀²|
(1/2)v₀²sin²θ = 7/16 * v₀₂
(1/2)sin²θ = 7/16
sin θ = [2*(7/16)]² = 0.765625
θ = sin⁻¹(0.765625) = 49.96°
Therefore, the launch angle is 49.96°.
Answer:
final velocity = 9.6157 m/sec
Explanation:
To solve this question, we will use one of the equations of motion which is:
v^2 = u^2 + 2as
where:
v is the final velocity that we want to get
u is the initial velocity = 3.75 m/sec
a is the acceleration due to gravity = 9.8 m/sec^2
s is the distance = 4 m
Substitute with the givens in the equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (3.75)^2 + 2(9.8)(4)
v^2 = 92.4625
v = 9.6157 m/sec
Hope this helps :)