A projectile is fired at an upward angle of 60 degrees with a speed of 100m/s. It lands on a plateau 150m higher. What is the projectile's speed the moment before it strikes the plateau?

I'm thinking to just use the kinematic equation Vf^2= Vi^2 + 2aΔy. Am I correct?

Vo = 100m/s[60o].

Xo = 100*Cos60 = 50 m/s.
Yo = 100*sin60 = 86.6 m/s.

Y^2 = Yo^2 + 2g*h = 86.6^2 - 19.6*150 =
4560.
Y = 67.5 m/s.

Well, you're definitely on the right track! But instead of using the kinematic equation Vf^2 = Vi^2 + 2aΔy, you should use the equation Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time.

In this case, because the projectile is fired at an upward angle, the only force acting on it is gravity, which causes it to accelerate downward. So, the acceleration (a) would be -9.8 m/s^2, since gravity's acceleration is always pointing down.

Now, to find the time it takes for the projectile to hit the plateau, we can use the equation Δy = Vi*t + (1/2)*a*t^2, where Δy is the change in vertical displacement (which is 150m in this case), Vi is the initial vertical velocity, t is the time, and a is the acceleration.

Using the information provided, we can calculate the initial vertical velocity by multiplying the initial speed of 100 m/s by the sine of the launch angle (60 degrees). So, Vi = 100 m/s * sin(60 degrees).

Now, you can rearrange the equation to solve for time (t). Once you have the value of t, you can substitute it back into the equation Vf = Vi + at to find the final velocity (Vf). And that should give you the projectile's speed the moment before it strikes the plateau.

I hope this helps, and good luck with your calculation!

Yes, you are correct! The kinematic equation you mentioned, Vf^2 = Vi^2 + 2aΔy, can be used to solve this problem.

In this case, the initial velocity (Vi) is given as 100 m/s, the angle of projection (θ) is 60 degrees, and the change in height (Δy) is 150 m (since the projectile lands on a plateau 150 m higher).

To use the equation, you'll need to determine the acceleration (a) in terms of the given values. In this case, the only force acting on the projectile in the vertical direction is gravity. So, the acceleration due to gravity (g) can be used as the vertical acceleration (a), with a value of approximately 9.8 m/s^2.

Now, substitute the values into the equation:

Vf^2 = Vi^2 + 2aΔy

Vf^2 = (100 m/s)^2 + 2 * (9.8 m/s^2) * 150 m

Vf^2 = 10000 m^2/s^2 + 2940 m^2/s^2

Vf^2 = 12940 m^2/s^2

Finally, take the square root of both sides to find the final velocity (Vf) just before the projectile strikes the plateau:

Vf = √(12940) ≈ 113.7 m/s

Therefore, the projectile's speed at the moment before it strikes the plateau is approximately 113.7 m/s.

Yes, you are correct. The kinematic equation Vf^2 = Vi^2 + 2aΔy is indeed the appropriate equation to use in this scenario. Let's break it down step by step to find the projectile's speed right before it strikes the plateau.

First, let's assign some values to the variables in the equation:
Vf = final velocity (the projectile's speed right before it strikes the plateau)
Vi = initial velocity (the projectile's speed when it was fired)
a = acceleration (in this case, it is the acceleration due to gravity which is approximately -9.8 m/s^2)
Δy = change in position in the vertical direction (in this case, it is the height difference between the projectile's launch point and the plateau, which is 150 m)

Now, let's plug in the values into the equation:
Vf^2 = Vi^2 + 2aΔy

Since the projectile was fired at an upward angle of 60 degrees with a speed of 100 m/s, we can break down the initial velocity into its horizontal and vertical components. The vertical component is given by Vi * sinθ, where θ is the angle of projection. So, Vi = 100 m/s * sin(60°) = 100 m/s * 0.866 ≈ 86.6 m/s.

Plugging in the values, we have:
Vf^2 = (86.6 m/s)^2 + 2 * (-9.8 m/s^2) * 150 m

Calculating this expression, we get:
Vf^2 ≈ 86.6^2 + (-9.8) * 300 ≈ 7510

Taking the square root of both sides of the equation, we find:
Vf ≈ √7510 ≈ 86.7 m/s

Therefore, the projectile's speed right before it strikes the plateau is approximately 86.7 m/s.

since the ball drop 100m above the ground,we calculate max height to get the vy at the point of landing:h=(100sin60)^2/20=375m so it travels(375m-100m) from the max height to the top of building vertically:v^2=2gh=2*10*275=5500 v=74.1m/s since the vx=100cos60=50m/s is constant,so it land with resultant velocity of:sqrt(74.1^2+50^2)=89.4m/s