A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 105 m/s at an angle of 37.0o with the horizontal, as shown in Figure 3-39. (a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the range X of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point P, find (c) the horizontal and vertical components of the velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal.
How do you do part C
Part C
Vo = 105m/s theta= 37.0 degrees
Horizontal = 105*cos37 = 83.86m/s
Vertical = 105*sin37 = 63.19m/s
To answer these questions, we'll break down the problem step by step. Let's start with part (a).
(a) Determine the time taken by the projectile to hit point P at ground level:
To solve for time, we can use the kinematic equation for vertical displacement:
y = y0 + v0*sin(θ)*t - (1/2)g*t^2
where:
y = vertical displacement (0 since it hits the ground at point P)
y0 = initial vertical position (125 m)
v0 = initial speed (105 m/s)
θ = launch angle (37.0°)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Plugging in the known values:
0 = 125 + 105*sin(37.0°)*t - (1/2)*9.8*t^2
Simplifying the equation:
-4.9t^2 + 105*sin(37.0°)*t + 125 = 0
This is a quadratic equation in terms of time (t). Solving it will give us the time taken by the projectile to hit point P.
Now, let's move on to part (b).
(b) Determine the range (X) of the projectile as measured from the base of the cliff:
The range of a projectile can be calculated using the horizontal component of its initial velocity and the time it takes to reach the ground.
To find the horizontal component of the initial velocity (v0x):
v0x = v0*cos(θ)
Plugging in the known values:
v0x = 105*cos(37.0°)
To find the range (X), we multiply the horizontal component of initial velocity (v0x) by the time (t):
X = v0x * t
We already found v0x in the previous step, and we'll use the value of t we find from the quadratic equation to calculate the range.
Moving on to part (c).
(c) Determine the horizontal and vertical components of the velocity just before the projectile hits point P:
Just before the projectile hits point P, its vertical velocity will be entirely downward, and its horizontal velocity will remain constant. Therefore, the horizontal component of the velocity will be the same as the initial horizontal velocity (v0x), and the vertical component will be the negative of the initial vertical velocity.
For the horizontal component: v_x = v0x
For the vertical component: v_y = -v0*sin(θ)
Let's move on to part (d).
(d) Determine the magnitude of the velocity just before the projectile hits point P:
The magnitude of the velocity can be calculated using the horizontal and vertical components:
|v| = sqrt(v_x^2 + v_y^2)
Plugging in the values we found for v_x and v_y, we can calculate the magnitude of the velocity.
Finally, let's tackle part (e).
(e) Determine the angle made by the velocity vector with the horizontal just before the projectile hits point P:
We can find this angle using trigonometry. The angle is given by:
θ' = tan^(-1)(v_y / v_x)
Plugging in the values we found for v_x and v_y, we can calculate the angle.
These steps provide a systematic approach to solving the problem. Plug in the known values and solve each step using the appropriate equations to find the answers to the given questions.
Vo = 105 m/s @ 37 Deg.
Xo = hor. = 105*cos37 = 83.9 m/s.
Yo = ver. = 105*sin37 = 63.2 m/s.
h = ho + (Y^2-Yo^2)/2g.
h = 125 + (0-(63.2)^2) / -19.6=328.8 m
above gnd.
Tr = (Y-Yo)/g=(0-63.2) / -9.8=6.45 s.
= Rise time.
a. h = Vo*t + 4.9t^2 = 328.8 m.
0 + 4.9t^2 = 328.8.
t^2 = 328.8 / 4.9 = 67.1.
Tf = 8.19 s. = Fall time.
b. Range=Xo*(Tr+Tf)=83.9(6.45+8.19)=1228.3 m.
c. Y = Yo + gt.
Y = ver. = 0 + 9.8*8.19 = 80.3 m/s.
X = hor = Xo = 83.9 m/s.
d. V = sqrt(X^2+Y^2).
V = sqrt((83.9)^2+(80.3)^2) = 116 m/s.
e. tanA = Y/X = 80.3 / 83.9 = 0.95709.
A = 43.7 Deg.