A projectile is fired with an initial speed of 120m.s' at an angle of 60' above the horizontal from the top of a 50m high cliff. (a) determine the maximum height(above the cliff) reached by the projectile (b) how long it takes to fall to the ground from the max. height?. Please give me some help i'm confuse.
use formula vfy= viy+gt...
vfy= 0, this is always true when the object reaches its maximum height.
viy= vi sin degress (horizontal)
viy= 120sin 60
viy= 103.92
g= -9.81 m/s2
thus:
vfy= viy + gt
0= 103.92 + (-9.81)(t)
t = -1o3.92/-9.81
1o.59seconds from the top of the cliff.
getting the time...
compute for the dy or the height from the top of the cliff...
dy= viyt + 1/2gt (time is squared)
add then the dy from the top of the cliff with the height of your cliff which is 5o to get the max.height.
using the new dy, (50 + the dy you computed from the top of the cliff), use the same formula of dy... now the viy is 0 since when the object is already falling, the final speed when the object has reached it max height is o, going down, the value vf will become the vi of the initial speed.
(im not sure but i guess this is how it is done)
the get the distance of the flight by:
dy=viyt - 1/2 gt2 (please square the t,cant do superscript here). Suppose you get the distance
To solve this problem, we can use the equations of motion for projectile motion.
(a) To determine the maximum height reached by the projectile, we need to find the vertical component of the initial velocity. The initial velocity can be broken down into horizontal and vertical components using trigonometry.
Given:
Initial speed (v): 120 m/s
Launch angle (θ): 60 degrees
Height of the cliff (h): 50 m
Vertical component of initial velocity (v₀y):
v₀y = v * sin(θ)
Using the given values:
v₀y = 120 * sin(60)
v₀y = 103.92 m/s (rounded to two decimal places)
The maximum height can be found using the equation:
h = (v₀y^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values, we have:
h = (103.92^2) / (2 * 9.8)
h ≈ 539.04 meters (rounded to two decimal places)
Therefore, the maximum height reached by the projectile is approximately 539.04 meters above the cliff.
(b) To determine how long it takes to fall to the ground from the maximum height, we need to find the time taken for the projectile to reach the maximum height and then double it.
The time taken to reach the maximum height can be found using the equation:
v = v₀y - g * t
Since the projectile is at its maximum height, the vertical component of the velocity is zero (v = 0). Rearranging the equation, we have:
t = v₀y / g
Substituting the values:
t = 103.92 / 9.8
t ≈ 10.61 seconds (rounded to two decimal places)
Since the time taken to reach the maximum height is the same as the time taken to fall back to the ground (neglecting air resistance), the total time taken to fall from the maximum height is twice the time taken to reach the maximum height:
Total time taken = 2 * t ≈ 2 * 10.61 s ≈ 21.22 seconds
Therefore, it takes approximately 21.22 seconds to fall to the ground from the maximum height.