a stone propelled from a catapult with a speed of 50mls,attains a height of 100m. Calculate the angle of projection and range.

HENRY, it's a good job.

To calculate the angle of projection and range of a stone propelled from a catapult, we can use the equations of projectile motion.

First, let's determine the initial vertical and horizontal velocities of the stone.

Given:
Initial speed (u) = 50 m/s
Initial vertical displacement (y) = 100 m

We know that the maximum height is attained at the time when the vertical velocity becomes zero (v = 0).

Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement, we can determine the time taken to reach the maximum height.

At the maximum height, the final vertical velocity (v) is 0 m/s, acceleration due to gravity (g) is -9.8 m/s^2, and the initial displacement (y) is 100 m.

0^2 = 50^2 + 2(-9.8)s
0 = 2500 - 19.6s
19.6s = 2500
s = 2500 / 19.6
s ≈ 127.55 m

Therefore, the stone reaches the maximum height at approximately 127.55 m.

Now we can calculate the time taken (t) to reach the maximum height using the equation v = u + at.

0 = 50 - 9.8t
9.8t = 50
t = 50 / 9.8
t ≈ 5.10 s

Since the time taken to reach the maximum height is half the total time of flight (T), we can calculate T using the equation T = 2t.

T = 2(5.10)
T = 10.2 s

Now, let's calculate the range (R) using the equation R = u * cos(theta) * T.

Given:
Initial speed (u) = 50 m/s
Total time of flight (T) ≈ 10.2 s

We need to find the angle of projection (theta) that will give us the range. Rearranging the formula, we get:

R / (u * T) = cos(theta)

Now, substitute the values:

R / (50 * 10.2) = cos(theta)

To find the value of cos(theta), divide the range (R) by the product of initial speed (u) and time of flight (T):

cos(theta) = R / (50 * 10.2)

Finally, find the angle of projection (theta) by taking the inverse cosine (cos^(-1)) of (R / (50 * 10.2)) using a calculator.

Once you have the value of cos^(-1)(R / (50 * 10.2)), you will obtain the angle of projection.

Vo = 50m/s @ A Degrees

Xo = 50*Cos A
Yo = 50*sin A

Y^2 = Yo^2 + 2g*h = 0 @ max ht.
Yo^2 = -2g*h = -2*(-9.8)*100 = 1960
Yo = 44.27 m/s = Ver. component of initial velocity.

Yo = 50*sin A = 44.27
sin A = 44.27/50 = 0.88544
A = 62.3o. = Angle of projection.

Y = Yo + g*Tr = 0
Tr = -Yo/g = 4.52 s = Rise time.

Tf = Tr = 4.52 s. = Fall time.

Range = Xo*(Tr+Tf) = 50*Cos62.3*(9.04) =
210 m.