Physics Help! Thanks! Motion Problem.?

Question

I need help with how to do with problem. If someone can show me the steps. I really don't get this.

An object is launched at an angle of 20 degrees with a speed of 30 m/s. It hits 4.0m up the face of a wall. How far away is the wall? (Note: Use Quadratic formula to find time.)

Answer 1

First we will have a horizontal and vertical components
Vv=Vsin(30)
Vh=Vcos(30)

Let the distance to the wall be

S=Vh t or S=Vcos(30) t

since h=4m and
h=Vv t - 0.5 gt^2

time t can be found by solving


0.5 gt^2 -Vv t +h=0
t^2 - (2/g)Vv t + (2/g)h=0

so "Use Quadratic formula to find time"

Answer 2

Here's a hint: since this is a simple trajectory problem, the path of the projectile is symmetrical and there are two answers. It can either hit the wall at the same height on its way up or on its way down as it follows its parabolic path in our gravitational field.

A projectile has both vertical and horizontal components to its velocity. The vertical is found by multiplying the initial velocity by the sine of the launch angle and changes with time, while the horizontal is the initial velocity times the cosine of the launch angle and is constant throughout the flight.

In this case these two velocities are 10.26 and 28.19 m/s respectively (keeping a few extra decimals because they are intermediate results).

The initial vertical velocity equals the final vertical velocity in the opposite direction, so we can calculate how long the flight lasts with the formula v=at and solve for t = v / a = 1.05 seconds for each half of the flight.

Now use distance = 1/2 a t^2 = 4.9 * 1.05^2 = 5.40 meters for the maximum height (and the maximum horizontal distance traveled is 28.19 * 1.05 * 2 = 59.2 meters).

Since the projectile hits the wall at a height of 4.0 meters, all we need to do next is subtract this from 5.40 and find how long it would take for something to fall the difference of 1.4 meters. Use the distance formula again and solve for time: 1.4 = 1/2 * a * t^2, so t^2 = 1.4 / 4.9 = 0.2857, and t = 0.53 seconds.

In other words, it takes 1.05-0.53=0.52 seconds for the object to reach a height of 4.0 meters (remember, the velocity is faster when it starts out). Multiply this by the constant horizontal velocity of 28.19 m/s to get about 14.7 meters for the distance of the wall from either the beginning or the end of the trajectory.