Take g= 10 MS^-2. A stone propelled from a catapult with a speed of 50 MS^-1 attains a height of 100 m. Calculate:

Question

a. the time of flight.
b. the angle of projection.
c. the range attained.

Answer 1

To solve this problem, we can use the equations of projectile motion. Let's use the following variables:

v = initial velocity of the stone (50 m/s)
g = acceleration due to gravity (-10 m/s^2, since it's acting against the motion of the stone)
h = maximum height of the stone (100 m)

a. To find the time of flight, we can use the equation for vertical motion:

h = (v^2 * sin^2θ) / (2g)

Rearranging the equation to solve for time, t:

2gh = v^2 * sin^2θ
t = 2v * sinθ / g

t = 2 * 50 * sinθ / (-10)
t = -10 * sinθ

We know that t > 0, so we take the positive value of sinθ:

t ≈ 10 seconds

b. To find the angle of projection, we can use the equation for vertical motion:

v = v₀ * sinθ - g * t

Since the stone has no vertical velocity at the peak of its trajectory, we know that sinθ = 0 at that point. Thus:

0 = v₀ * sinθ - g * t
v₀ * sinθ = g * t

Substituting the known values:

50 * sinθ = 10 * 10
sinθ = 1

θ = arcsin(1)
θ ≈ 90 degrees

Therefore, the angle of projection is approximately 90 degrees.

c. To find the range attained, we can use the equation for horizontal motion:

R = v₀ * cosθ * t

Substituting the known values:

R = 50 * cos90 * 10
R = 0

Therefore, the range attained is 0 meters.