A projectile is shot from the edge of a cliff h = 285 m above ground level with an initial speed of v0 = 155 m/s at an angle of 37.0° with the horizontal.
(a) Determine the time taken by the projectile to hit point P at ground level.
Incorrect: Your answer is incorrect.
s
(b) Determine the range X of the projectile as measured from the base of the cliff.
km
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
(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.
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
m/s (horizontal component)
m/s (vertical component)
(d) What is the the magnitude of the velocity?
m/s
(e) What is the angle made by the velocity vector with the horizontal?
° (below the horizontal)
To solve this problem, we can break it into several parts. Let's start by calculating the time taken by the projectile to hit point P at ground level.
(a) To find the time taken, we need to consider the vertical motion of the projectile. We can use the equation:
h = (v0*sinθ)*t - (1/2)*g*t^2
Where:
h = height of the cliff (285 m)
v0 = initial speed (155 m/s)
θ = launch angle (37.0°)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken
Rearranging the equation, we get:
(1/2)*g*t^2 - (v0*sinθ)*t + h = 0
This is a quadratic equation in terms of t. We can solve it by using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Substituting the values, we have:
a = (1/2)*g = 4.9 m/s^2
b = - (v0*sinθ) = - (155 m/s * sin(37.0°))
c = h = 285 m
Now, we can plug in these values into the quadratic formula and solve for t.
(b) To determine the range X of the projectile as measured from the base of the cliff, we need to consider the horizontal motion of the projectile. The horizontal range can be given by:
X = (v0*cosθ) * t
Now that we have the value of t from part (a), we can calculate X.
(c) To find the horizontal and vertical components of velocity just before the projectile hits point P, we can use the equations:
Vx = v0*cosθ
Vy = v0*sinθ - g*t
Substituting the values, we can evaluate Vx and Vy.
(d) The magnitude of the velocity can be determined using the equation:
v = √(Vx^2 + Vy^2)
Substitute the values and calculate the magnitude of velocity.
(e) Finally, to find the angle made by the velocity vector with the horizontal, we can use the equation:
θ_velocity = arctan(Vy/Vx)
Substitute the values and calculate the angle.
By following these steps, you should be able to find the answers for parts (a) to (e).