A projectile is fired with an initial speed of 120m.s' at an angle of 60' above the horizontal from the top of a 50m high cliff. (a) determine the maximum height(above the cliff) reached by the projectile (b) how long it takes to fall to the ground from the max. height?. Please give me some help i'm confuse.
1st you find the vertical component of its speed which = (120)sin(60)=103.92 m/s
Gravity's pull will decrease its vertical speed by G (=9.8 m/s^2)
Therefore the vertical velocity of the projectile = 103.92 - (9.8)*t (whee t is time in seconds)
It will reach maximum height when the velocity is 0 (because then it will start falling)
So 103.92 - (9.8)*t = 0 when t = 10.6 seconds (maximum height is reached when t = 10.6 seconds)
to find the height we integrate the vertical velocity eq' (103.92 - (9.8)*t) and we get ( (103.92)*t - (4.9)*t^2 + C ) (note: Since at t = 0, the height = (103.92)*(0) - (4.9)*(0)^2 + C = 0, therefor C = 0. If we assume at t = 0 our height is 50 (because of the cliff, then C = 50 but here we will use the cliff as our reference point therefore C = 0)
we sub in t = 10.6 and we get the maximum height above the cliff = (103.92)*(10.6) - (4.9)*(10.6)^2 = 550.99 meters above the cliff (or 600.99 meters above ground)
Part b:
We can use the vertical velocity equation we used before (103.92)*t - (4.9)*t^2 + C
But here we will put C = 50 because the ground will be our reference point.
Therefore (103.92)*t - (4.9)*t^2 + 50 = 0
Solving this eq' we get t = -0.47 seconds or t = 21.68 seconds, since we cant have negative time we take t = 21.68 seconds
At max height t was 10.6 Seconds, Then time it took from max height to ground = 21.68 - 10.6 = 11.08 seconds
thanks a lot Nadir
Well, let me help you out by breaking it down with a little bit of humor!
(a) To determine the maximum height reached by the projectile, we need to consider the vertical motion. It's like watching a gymnast doing flips, but in this case, our projectile is the star of the show!
First, we can find the time it takes for the projectile to reach its highest point. It's like when your favorite ice cream falls off the cone — it reaches its highest point before the inevitable splat on the ground.
Using the initial speed (120 m/s) and the launch angle (60 degrees), we can find the time it takes to reach the highest point using some fancy shmancy trigonometry. The formula is t = (V0 * sinθ) / g, where V0 is the initial speed, θ is the launch angle, and g is the acceleration due to gravity.
Once you find the time, you can use it to find the maximum height above the cliff using the formula h = V0^2 * (sinθ)^2 / (2g). It's like reaching the top of a roller coaster and feeling the adrenaline pumping!
(b) Now, to determine how long it takes to fall to the ground from the maximum height, we need to consider the vertical motion once again. It's like waiting for the punchline of a joke — you know it's coming!
We can use the formula t = √(2h / g), where h is the maximum height we found earlier and g is the acceleration due to gravity. This formula gives us the time it takes for the projectile to fall back to the ground.
So, with a little bit of math and a sprinkle of humor, you'll be able to determine the maximum height reached by the projectile and the time it takes to fall back down. Good luck! And remember, laughter is the best projectile!
To solve this problem, we can break it down into two parts: the projectile's motion while it's going up, and its motion while it's coming down.
Let's start by solving part (a), which asks for the maximum height reached by the projectile.
First, we need to find the time it takes for the projectile to reach its maximum height. We can use the formula for the time of flight for a projectile:
time_of_flight = (2 * initial_velocity * sin(theta)) / gravity
where:
- initial_velocity = 120 m/s (initial speed of the projectile)
- theta = 60 degrees (angle of projection)
- gravity = 9.8 m/s^2 (acceleration due to gravity)
Plugging in the values, we get:
time_of_flight = (2 * 120 * sin(60)) / 9.8
Next, we can use the formula for the maximum height of a projectile:
maximum_height = (initial_velocity^2 * sin^2(theta)) / (2 * gravity)
Plugging in the values, we get:
maximum_height = (120^2 * sin^2(60)) / (2 * 9.8)
Now we can calculate the values:
time_of_flight = (2 * 120 * sin(60)) / 9.8 ≈ 14.4 seconds
maximum_height = (120^2 * sin^2(60)) / (2 * 9.8) ≈ 104.1 meters
So, the maximum height reached by the projectile above the cliff is approximately 104.1 meters.
Now, let's move on to part (b), which asks for the time it takes for the projectile to fall to the ground from the maximum height.
Since the time taken to reach the highest point is the same as the time taken to fall back to the ground, the answer is the same as the time of flight we calculated earlier, which is approximately 14.4 seconds.
Therefore, it takes approximately 14.4 seconds for the projectile to fall to the ground from the maximum height.
I hope this explanation helps! Let me know if you have any further questions.