A stone propelled from a catapult with a speed of 100m/s and attains a height of 50 m .What would be the horizontal distance traveled
Y^2 = Yo^2 + 2g*h = 0 m/s.
Yo^2 + (-19.6)50 = 0,
Yo = 31.3 m/s. = Ver. component of initial velocity.
Yo = Vo*sinA = 31.3.
100*sinA = 31.3,
sinA = 0.313,
A = 18.2o.
Xo = Vo*CosA = 100*Cos18.2 = 95 m/s. = Hor. component of Vo.
Y = Yo + g*t = 0.
31.3 + (-9.8)t = 0,
t = 3.19 s. = Rise time.
d = Xo * t = 95 * 3.19 = meters.
To find the horizontal distance traveled by the stone, we need to calculate the horizontal component of its motion.
First, let's consider the time it takes for the stone to reach its maximum height. We can use the formula for the vertical motion of an object under constant acceleration:
v_f^2 = v_i^2 + 2*a*d,
where:
v_f = final velocity (0 m/s at the maximum height),
v_i = initial velocity (100 m/s),
a = acceleration (acceleration due to gravity, which is approximately -9.8 m/s^2), and
d = displacement (height, 50 m).
Plugging in the values, we get:
0^2 = (100^2) + 2*(-9.8)*50.
Simplifying this equation, we find:
0 = 10000 - 980.
Rearranging and solving for t, we get:
t = sqrt(10000 / 980).
t ≈ 3.19 seconds.
Now, since the motion is symmetrical, it takes the same amount of time for the stone to fall from its maximum height to the ground. Therefore, the total time of flight is 2 * t.
Using the formula for horizontal motion:
d = v * t,
where:
d = horizontal distance traveled (what we want to find),
v = horizontal velocity (which is constant and equal to the initial velocity, since there is no horizontal acceleration),
t = total time of flight (2 * t),
we can calculate the horizontal distance traveled by the stone.
Plugging in the values, we get:
d = 100 * 2 * t.
Substituting the value of t we found earlier, we have:
d ≈ 100 * 2 * 3.19.
Calculating this, we get:
d ≈ 638 meters.
Therefore, the horizontal distance traveled by the stone would be approximately 638 meters.