Projectile motion problem (on the moon)

[highwind7777]

actually, i have 2 problems I'm having terrible problems with:

#1: i throw a ball with an intial SPEED of 15m/s at an unknown angle. there's a 2metre high wall 25 metres in front of me. between which 2 angles should i aim the ball for it to fly above this wall?

#2: i throw a ball at unknown speed at an inclination of 30 degrees, and it goes over a fence (of unknown height...) 80 metres away, and it's total time of flight is 4.72 seconds (also note that the ball does not necessarily end up on the same ground level as where i threw it from. it could have gone under a cliff, or up a hill...) if this exact scenario took place on the moon (which has gravity of 1.627m/s^2) instead of on earth, what would the time of flight be?

 

[Leong]

#1

Assume it occurred on the moon.
x-component:
s=ut
$$25=(15cos\theta)t$$…(1)
$$t=\frac{5}{3cos\theta}$$
y-component:
$$s=ut-\frac{g}{2}t^2>2$$…(2)
(1) into (2) and simplify with g=1.627 meter per second per second:
$$225cos\theta sin\theta-20.3375>18cos^2\theta$$…(3)
$$Let\ cos\theta=x;\ then\ sin\theta=\sqrt{1-x^2}$$and substitute into (3) and simplify :
$$x\sqrt{1-x^2}>0.08x^2+0.09039$$
Square both sides and simplify :
$$1.0064x^4-0.98554x^2+0.008170<0$$
Find the roots using quadratic formula:
$$x=\pm 0.9854\ or\ \pm0.09144$$
Draw a line number and identify the area which satisfy the inequality :
$$0.09144<x<0.9854$$
$$0.09144<cos\theta<0.9854$$
$$cos^{-1}0.9854<\theta<cos^{-1}0.09144$$
$$9.80^0<\theta<84.7^0$$

 

[wais]

#1: To solve this problem, we can use the equations of projectile motion. First, we need to find the time it takes for the ball to reach the wall. Using the equation t = 2v*sinθ/g, where t is the time, v is the initial velocity, θ is the angle and g is the acceleration due to gravity (on the moon, g = 1.627m/s^2), we get t = 2*15*sinθ/1.627 = 9.23*sinθ seconds. Now, we can use the equation d = v*t*cosθ, where d is the distance, v is the initial velocity, t is the time and θ is the angle. We know that d = 25m and t = 9.23*sinθ seconds. Substituting these values, we get 25 = 15*9.23*sinθ*cosθ/1.627. Solving this equation, we get two possible angles: θ = 19.5 degrees and θ = 70.5 degrees. Therefore, to ensure that the ball flies above the wall, you should aim between these two angles.

#2: To find the time of flight on the moon, we can use the same equation t = 2v*sinθ/g, where t is the time, v is the initial velocity, θ is the angle and g is the acceleration due to gravity (on the moon, g = 1.627m/s^2). We know that t = 4.72 seconds, θ = 30 degrees and g = 1.627m/s^2. Substituting these values, we get 4.72 = 2v*sin30/1.627. Solving for v, we get v = 15.33m/s. Now, we can use the equation h = v^2*sin^2θ/2g, where h is the maximum height reached by the ball. We know that h = 80m, v = 15.33m/s, θ = 30 degrees and g = 1.627m/s^2. Substituting these values, we get 80 = 15.33^2*sin^2(30)/2*1.627. Solving this equation, we get h = 29.3m. Therefore, the total time of flight on the

 

[M98Ranger]

For problem #1, in order to find the necessary angles to aim the ball, we can use the equation for projectile motion, which is:

y = y0 + v0sinθt - 1/2gt^2

Where:
y = final height (in this case, 2m)
y0 = initial height (in this case, 0m)
v0 = initial velocity (in this case, 15m/s)
θ = angle of launch (unknown)
t = time of flight (unknown)
g = acceleration due to gravity on the moon (1.627m/s^2)

We can rearrange the equation to solve for θ:

θ = sin^-1 [(y-y0+1/2gt^2)/v0t]

Plugging in the values, we get:

θ = sin^-1 [(2-0+1/2(1.627)t^2)/(15t)]

To find the minimum and maximum angles, we can use the fact that the maximum height of the ball occurs when the vertical component of the initial velocity is 0, so:

v0sinθ = 0

Therefore, the minimum angle is 0 degrees, and the maximum angle is 90 degrees. So, you should aim the ball between 0 and 90 degrees to ensure it goes above the 2m wall.

For problem #2, we can use the same equation for projectile motion, but with the given values for the moon's gravity and the time of flight:

y = y0 + v0sinθt - 1/2gt^2

Where:
y = final height (unknown)
y0 = initial height (unknown)
v0 = initial velocity (unknown)
θ = angle of launch (30 degrees)
t = time of flight (4.72 seconds)
g = acceleration due to gravity on the moon (1.627m/s^2)

We can rearrange the equation to solve for y:

y = y0 + v0sinθt - 1/2gt^2

Plugging in the values, we get:

y = 0 + v0(0.5)(4.72) - 1/2(1.627)(4.72)^2

Solving for y, we get:

y = 0.5v0(4.72) - 3.84

Since we do not know the initial velocity or