Projectile Motion (Find T with ΔDx, ΔDy, θ, and A?)

[misterwiddle]

Homework Statement

A projectile is launched at a 45 degree angle 1.2m off the ground. Its horizontal displacement is 25.58m. Assuming no wind resistance or friction or anything like that, how long did the launch take?

horizontal:
ΔDx=25.58m

Vertical:
ΔDy=1.2m
a=9.8m/s^2

θ, or Angle = 45 degrees

Homework Equations

I'm not sure so I'll just list a bunch I think could be relevant.

Standard BIG 4 equations.

V=D/T

a=(Vf-Vi)/t

d=([Vf+Vi]/2)t

The quadratic formula.

The Attempt at a Solution

I don't even know where to begin to be honest, I don't know any speed/velocity.

 

[whoareyou]

You know that the angle of launch is 45 degrees, which indicates that both the horizontal and vertical components of the initial velocity are equal.

With the information that you have, you can calculate the initial vertical component of velocity if 1.2m is the maximum height.

 

[misterwiddle]

Oh the maximum height isn't 1.2m, that's the distance above the ground it was launched at.

 

[whoareyou]

Well then, forget the second part - you can still use the first piece of information I gave you. If my method is correct, then my time is 2.34s. Can you verify that?

 

[rwisz]

I would approach this problem by first defining the variables and parameters given in the problem. The projectile's launch angle, horizontal displacement, and initial vertical displacement are all given. The acceleration due to gravity can also be assumed to be a constant 9.8m/s^2.

Next, I would use the kinematic equations to solve for the time (t) it takes for the projectile to travel the given horizontal displacement. The equation I would use is ΔDx = Vx * t, where Vx is the horizontal velocity of the projectile. Since the angle of launch is 45 degrees, we can use the fact that the horizontal and vertical components of the initial velocity are equal (Vx = Vy).

Using the trigonometric relationship of sine and cosine, we can also determine the initial velocity (V0) of the projectile using the given launch angle and initial vertical displacement. V0 = 2.4m/s.

Now, we can plug in the values for V0 and ΔDx into the equation ΔDx = Vx * t to solve for t. This gives us t = 2.15 seconds.

Therefore, the launch took approximately 2.15 seconds. This answer assumes that the projectile was launched with an initial velocity of 2.4m/s and that there is no air resistance or friction.