A students on a balcony uses a slingshot to shoots rock straight up
at 29.4 m/s. If the rock takes 8.0 s seconds to hit the ground below the balcony, how high is the balcony above the ground?
The velocity 0f the rock is zero at its' max. height:
Vf = Vo + gt = 0,
29.4 - 9.8t = 0,
-9.8t = -29.4,
t = -29.4 / -9.8 = 3s to reach max height
Time in flight = 8s.
Fall time = 8 - 3 = 5s,
d = 0*5 + 0.5*9.8*5^2,
d = 0 + 122.5 = 122.5m to gnd.
h = 122.5 - 44.1 = 78.4m.
= 4.9*3^2,
d = 88.2 - 44.1 = 44.1m up.
Oops!!
d (up) = Vo*t + 0.5*gt^2,
d (up) = 29.4*3 + 0.5*(-9.8)3^2,
d (up) = 88.2 - 44.1 = 44.1m.
To determine the height of the balcony above the ground, we can use the principles of projectile motion. First, let's break down the problem into its key components.
Given:
Initial velocity (u) = 29.4 m/s (the rock is shot straight up)
Time of flight (t) = 8.0 s (the rock takes this long to hit the ground)
Acceleration due to gravity (g) = 9.8 m/s^2 (assuming no air resistance)
We need to find the height of the balcony above the ground (h). To do this, we can use the equation of motion:
h = u * t - (1/2) * g * t^2
Substituting the given values:
h = (29.4 m/s) * (8.0 s) - (1/2) * (9.8 m/s^2) * (8.0 s)^2
Calculating:
h = 235.2 m - (1/2) * 9.8 m/s^2 * 64.0 s^2
h = 235.2 m - 313.6 m
h = -78.4 m
The result is -78.4 m, which indicates that the height of the balcony is negative. This means the calculation assumes that the ground is at a higher elevation than the balcony, which is unlikely. Therefore, please check if the initial velocity or the time of flight is correctly provided.