A projectile is shot from the edge of a cliff h = 225 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.
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.)
horizontal:
Vertical:
To solve this problem, we can use the equations of motion for projectile motion.
a. Determine the time taken by the projectile to hit point P at ground level:
The vertical motion of the projectile can be described using the equation h = h0 + v0y * t - (1/2) * g * t^2, where h0 is the initial height of the projectile, v0y is the vertical component of the initial velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken. Since the projectile is at the level of the cliff when it is shot, h0 = 0.
We know that the vertical component of the initial velocity is given by v0y = v0 * sin(θ), where θ is the angle with respect to the horizontal. Therefore, v0y = 155 m/s * sin(37.0°).
Solving the equation for t, we get: 225 m = 0 + (155 m/s * sin(37.0°)) * t - (1/2) * (9.8 m/s^2) * t^2.
This is a quadratic equation, which we can solve using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = (-1/2) * (9.8 m/s^2), b = (155 m/s * sin(37.0°)), and c = -225 m.
Solving for t, we get two values. Since we only want the time it takes for the projectile to hit the ground (positive time value), we take the positive value.
b. Determine the range X of the projectile as measured from the base of the cliff:
The horizontal motion of the projectile can be described using the equation x = x0 + v0x * t, where x0 is the initial horizontal position, v0x is the horizontal component of the initial velocity, and t is the time taken.
We can calculate the horizontal component of the initial velocity using v0x = v0 * cos(θ), where θ is the angle with respect to the horizontal. Therefore, v0x = 155 m/s * cos(37.0°).
Substituting the values into the equation, we get x = 0 + (155 m/s * cos(37.0°)) * t.
c. At the instant just before the projectile hits point P, find the horizontal and vertical components of its velocity:
At the instant before hitting point P, the vertical component of the velocity is equal to the negative of the initial vertical component (v0y) because the projectile is moving opposite to the positive y-direction.
The horizontal component of the velocity remains the same throughout the motion because there is no horizontal acceleration.
Therefore, the horizontal component of the velocity is v0x = 155 m/s * cos(37.0°), and the vertical component of the velocity is -v0y = -155 m/s * sin(37.0°).
Thus, the solutions to the problem are:
a. The time taken by the projectile to hit point P at ground level is the positive value obtained from solving the quadratic equation.
b. The range X of the projectile is given by x = (155 m/s * cos(37.0°)) * t, where t is the time obtained in part a.
c. The horizontal component of the velocity is v0x = 155 m/s * cos(37.0°), and the vertical component of the velocity is -v0y = -155 m/s * sin(37.0°) just before the projectile hits point P.
To solve this problem, we can use the equations of motion for projectile motion. Let's break the problem down step by step:
Step 1: Find the time taken by the projectile to hit point P at ground level.
To find the time of flight, we can use the equation:
h = v0y * t - (1/2) * g * t^2
where:
h = height of the cliff = 225 m
v0y = vertical component of the initial velocity = v0 * sin(theta)
t = time of flight
g = acceleration due to gravity = 9.8 m/s^2
theta = angle with the horizontal = 37.0°
Plugging in the given values, we have:
225 = (155 * sin(37.0°)) * t - (1/2) * (9.8) * t^2
Simplifying the equation, we get a quadratic equation:
-4.9t^2 + 62.88t - 225 = 0
Using the quadratic formula, we can solve for t:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where:
a = -4.9
b = 62.88
c = -225
Solving the quadratic equation, we get two possible values of t: t = 5.59 s or t = 8.35 s.
Since we are interested in the time taken by the projectile to hit the ground, we take the positive value: t = 5.59 s.
Therefore, the time taken by the projectile to hit point P is 5.59 seconds.
Step 2: Find the range X of the projectile as measured from the base of the cliff.
To find the range, we can use the equation:
X = v0x * t
where:
X = range of the projectile
v0x = horizontal component of the initial velocity = v0 * cos(theta)
t = time of flight (which we found in Step 1)
Plugging in the given values, we have:
X = (155 * cos(37.0°)) * 5.59
Calculating this, we find that the range of the projectile is approximately X = 1037.46 m.
Therefore, the range of the projectile as measured from the base of the cliff is approximately 1037.46 meters.
Step 3: Find the horizontal and vertical components of the velocity just before the projectile hits point P.
At the instant just before the projectile hits point P, the vertical component of the velocity will be equal to zero, and the horizontal component of the velocity will remain the same as the initial horizontal velocity.
Therefore, the horizontal component of the velocity just before the projectile hits point P is v0x = 155 * cos(37.0°) = 123.81 m/s (to the right).
The vertical component of the velocity just before the projectile hits point P is zero.
Therefore, the horizontal component of the velocity is 123.81 m/s (to the right), and the vertical component of the velocity is zero.