A projectile is shot horizontally at 220 m/s from the top of a 620 m high cliff.
A) How long till the projectile hits the ground?
B) How far from the base of the cliff does the projectile land?
I am gonna be honest. I have no idea how to do any of my Physics homework. ): I still have 3 more questions to do too outta 5.
Split this into two problems. Since the projectile starts horizontal it keeps its horizontal speed. There is no horizontal force to change that speed.
On the other hand it starts with no vertical speed but there is a vertical force, gravity. Therefore it accelerates down at 9.8 m/s^2
A) is just a falling problem
h = (1/2) g t^2
620 = 4.9 t^2
t = 11.25 seconds
B) It goes for 11.25 seconds at 220 m/s
11.25*220 = 2475
thank you soooooo much(:
No worries! I'm here to help you with your physics homework. Let's start by tackling each question step by step.
A) To find how long it takes for the projectile to hit the ground, we can use the equation of motion for a falling object in the vertical direction:
y = v0t + (1/2)gt^2
Where:
- y is the vertical displacement (in this case, the height of the cliff, which is -620 m since it is downward)
- v0 is the initial vertical velocity (which is 0 m/s since the projectile is shot horizontally)
- t is the time taken to hit the ground (what we want to find)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the projectile is shot horizontally, its initial vertical velocity is zero, and the equation reduces to:
y = (1/2)gt^2
Substituting the given values:
-620 m = (1/2) * 9.8 m/s^2 * t^2
Solving for t, we get:
t^2 = (-620 m) / ((1/2) * 9.8 m/s^2)
t^2 = (-620 m) / 4.9 m/s^2
t^2 = 126.53 s^2
t ≈ √126.53 s
t ≈ 11.25 s
Therefore, it takes approximately 11.25 seconds for the projectile to hit the ground.
B) To find how far from the base of the cliff the projectile lands, we can use the equation of motion for a horizontally moving object:
x = v0x * t
Where:
- x is the horizontal displacement (what we want to find)
- v0x is the initial horizontal velocity (which is given as 220 m/s)
- t is the time taken to hit the ground (which we found in part A)
Substituting the given values:
x = 220 m/s * 11.25 s
x ≈ 2475 m
Therefore, the projectile lands approximately 2475 meters away from the base of the cliff.
Remember, it's important to understand the concepts and formulas in physics so you can apply them to solve problems effectively.