Velocity of 1-dimensional projectile

• notReallyHere
In summary, the projectile launched straight upward at +63.0 m/s from a height of 85.0 m, falls and just misses a sheer cliff before hitting the ground below. The maximum height reached is 202.5m. To find the time it takes to hit the ground at the base of the cliff, the equation xf=xi+vit+1/2at2 needs to be used, taking into account the initial velocity and acceleration due to gravity. The correct answer is found to be t= 6.429 seconds. To find the velocity at impact, the equation vf=vi+at can be used, but this depends on the correct answer to #1.
notReallyHere

Homework Statement

A projectile is launched straight upward at +63.0 m/s from a height of 85.0 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.

Maximum height = 202.5m

1) Find the time it takes to hit the ground at the base of the cliff.
2) Find its velocity at impact

xf=xi+vit+1/2at2
vf=vi+at

The Attempt at a Solution

For #1 I tried this:
xi=202.5 xf=0 a=-9.80 vi=0
0=202.5+0+1/2(-9.8)t2
-202.5=-4.9t2
41.327=t2
t=6.429

But that's not the right answer

For #2 I tried this: (It depends on the answer to #1)
Vf=0+(-9.8)(6.429)
Vf=-63.004

But that's wrong too since #1 is wrong. What did I do wrong in #1?

By putting x_i = 202.5, you are only solving for the time it takes to fall from the top of its motion, down to the ground. You have to account for the time it took to get up there!

Your mistake in #1 is that you have used the wrong initial position (xi) and final position (xf) values. The initial position should be 85.0 m (the height at which the projectile is launched from) and the final position should be 0 m (the ground level where the projectile hits). So the correct equation would be:

xf = xi + vit + 1/2at^2

0 = 85 + 63t + 1/2(-9.8)t^2

-85 = 63t - 4.9t^2

4.9t^2 - 63t - 85 = 0

Solving this quadratic equation, we get two values for t: t = 6.429 s (which is the time it takes for the projectile to reach its maximum height) and t = 14.189 s (which is the time it takes for the projectile to hit the ground).

For #2, at the time of impact (t = 14.189 s), the velocity of the projectile can be found using the equation:

vf = vi + at

vf = 63 + (-9.8)(14.189)

vf = -63.004 m/s

Note that the negative sign indicates that the velocity is in the downward direction, which makes sense since the projectile is falling.

1. What is the definition of velocity?

Velocity is a measure of an object's change in position over time. It is a vector quantity, meaning it has both magnitude (speed) and direction. In one-dimensional motion, the direction of velocity is typically denoted as positive or negative, depending on the direction of motion.

2. How is velocity calculated?

The mathematical equation for velocity is v = d/t, where v is velocity, d is displacement (change in position), and t is time. This means that velocity is equal to the distance traveled divided by the time it takes to travel that distance. The unit for velocity is typically meters per second (m/s).

3. What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement divided by the total time traveled. It gives an overall picture of an object's motion. Instantaneous velocity, on the other hand, is the velocity at a specific moment in time. It is calculated by taking the derivative of the object's position with respect to time.

4. How does velocity affect the trajectory of a projectile?

The velocity of a projectile plays a crucial role in determining its trajectory. The initial velocity and angle of launch determine the shape and distance of the projectile's path. The higher the velocity, the greater the horizontal displacement and the longer the projectile stays in the air.

5. Can velocity change during a projectile's flight?

Yes, velocity can change during a projectile's flight. This can happen due to factors such as air resistance, gravity, or external forces acting on the object. In a perfect vacuum with no external forces, the velocity of a projectile would remain constant throughout its flight.

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