A projectile leaves the top of a tower with a horizontal velocity of 13 m/s. Determine its horizontal displacement and vertical displacement if it lands at 70° from horizontal.
Guide On Rating System
Vote
The horizontal displacement (x) and vertical displacement (y) can be determined using the equations of motion.
First, we need to find the time of flight (t). We know that the angle at which the projectile lands is 70° from the horizontal. This means that the vertical component of the velocity when it lands is greater than the horizontal component.
We can use the equation v = u + gt to find the time of flight, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
In this case, the vertical component of the velocity when the projectile lands is v = 13 * tan(70°), the initial vertical velocity is u = 0 (since the projectile is launched horizontally), and g = 9.8 m/s².
So,
13 * tan(70°) = 0 + 9.8 * t
Solving for t gives t = 13 * tan(70°) / 9.8 ≈ 2.42 seconds.
Now we can find the horizontal and vertical displacements.
The horizontal displacement is given by x = ut + 0.5 * a * t², where u is the initial horizontal velocity, a is the horizontal acceleration, and t is the time.
Since the projectile is launched horizontally, there is no horizontal acceleration, so a = 0.
So,
x = 13 * 2.42 + 0.5 * 0 * (2.42)² = 31.46 meters.
The vertical displacement is given by y = ut + 0.5 * a * t², where u is the initial vertical velocity, a is the vertical acceleration, and t is the time.
In this case, u = 0 and a = -9.8 m/s² (since the acceleration is downwards).
So,
y = 0 * 2.42 + 0.5 * -9.8 * (2.42)² = -28.6 meters.
So, the horizontal displacement is 31.46 meters and the vertical displacement is -28.6 meters (the negative sign indicates that the displacement is downwards).
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99