A projectile is shot from the edge of a cliff h = 125 m above ground level with an initial speed of v0 = 115 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 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.
(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).
(d) What is the the magnitude of the velocity?
(e) What is the angle made by the velocity vector with the horizontal?
We don't get to see the figure 3-39, but I assume that the angle 37° is an angle of elevation, that is, it is shot upwards.
Resolve the velocity into the horizontal (cos37°) and vertical (sin37°) components.
Without air resistance, the horizontal component will not vary, i.e. it is a constant velocity.
The vertical component can be treated like a projectile shot vertically upwards, and the usual kinematics formulae will apply.
v0=115sin37° m/s
a=-9.8 m/s/s
S= -125 m when it hits the ground
t=time in seconds
2aS = v0t+(1/2)at²
Solve for the quadratic to get t.
Can you take it from here?
stuff
y 2aS = v0t+(1/2)at² ?????
I need help with the same problem! :(
To solve this problem, we can use the equations of projectile motion. Let's break down each part of the problem and go through the steps to find the answers.
(a) Time taken by the projectile to hit point P at ground level:
To find the time taken, we need to find the time it takes for the projectile to reach its highest point (when the vertical component of velocity is zero) and then double that time.
1. Find the time taken to reach the highest point of the projectile's trajectory:
Using the equation v = u + at, where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken, we can rearrange the equation to solve for t.
v = u + at
0 = 115sin(37) - 9.8t
9.8t = 115sin(37)
t = (115sin(37)) / 9.8
2. Double the time taken to reach the highest point to find the total time taken to hit point P at ground level:
Since the time it takes to reach the highest point is the same as the time taken to hit point P when the projectile comes back to the same height, we can double the time to find the total time taken. Multiply the result from step 1 by 2.
Total time taken = 2 * t
(b) Range X of the projectile as measured from the base of the cliff:
To find the range, we need to find the horizontal distance covered by the projectile.
1. Find the horizontal component of the initial velocity:
The horizontal component of the velocity can be found using the equation v = u + at, where v is the final horizontal velocity, u is the initial horizontal velocity, a is the horizontal acceleration (which is zero for a projectile in the absence of air resistance), and t is the time taken. Since there is no horizontal acceleration, the equation becomes:
v = u
v_horizontal = v0 * cos(37)
2. Find the total time for the projectile to hit point P:
We already calculated the total time taken in part (a).
3. Calculate the range using the formula:
Range = v_horizontal * total time taken
(c) Horizontal and vertical components of velocity just before hitting point P:
To find the components of velocity just before hitting point P, we can use the known values of initial velocity, angle, and time taken to find the final velocities in the horizontal and vertical directions.
1. Find the final vertical velocity:
Using the equation v = u + at, where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the total time taken, we can rearrange the equation to solve for v.
v = u + at
v_vertical = 115sin(37) - 9.8 * total time taken
2. Find the final horizontal velocity:
The horizontal component of velocity doesn't change throughout the motion because there is no horizontal acceleration. Therefore, the final horizontal velocity is the same as the initial horizontal velocity.
v_horizontal = v0 * cos(37)
(d) Magnitude of the velocity:
To find the magnitude of the velocity just before hitting point P, we can use the Pythagorean theorem with the horizontal and vertical components of the velocity.
Magnitude of velocity = square root of (v_horizontal^2 + v_vertical^2)
(e) Angle made by the velocity vector with the horizontal:
To find the angle made by the velocity vector with the horizontal just before hitting point P, we can use trigonometry to determine the angle whose tangent is the ratio of the vertical component to the horizontal component.
Angle = arctan(v_vertical / v_horizontal)
By following these steps and plugging in the given values, you should be able to find the answers to all parts of the problem.