a boy uses a slingshot to launch a rock with a velocity of 30m/s how long does it take it to reach a height of 20m,what maximum height does the rock reach, what is the velocity of the rock at this height, what is the velocity of the rock just before it lands
Vo = 30m/s[90o]?.
a. Vo*t + 0.5g*t^2 = 20 m.
30*t- 4.9t^2 = 20, -4.9t^2 + 30t - 20 = 0, t = ?. Use Quad. Eq.
b. V^2 = Vo^2 + 2g*h = 0. 30^2 - 19.6h = 0, h = ?.
c. V = 0 at max. ht.
d. V = Vo = 30 m/s.
To find the time it takes for the rock to reach a height of 20m, we need to use the kinematic equation:
h = v₀t + (1/2)gt²
Where:
h = height (20m)
v₀ = initial velocity (30m/s)
t = time
g = acceleration due to gravity (assuming -9.8 m/s²)
Rearranging the equation to solve for time, we have:
t = √(2h/g)
Substituting the known values, we get:
t = √(2 * 20 / -9.8)
t = √(40 / -9.8)
t = √(-4.08)
Since time can't be negative, we know that the rock will not reach a height of 20m and will not land back on the ground.
To find the maximum height the rock reaches, we can use the equation for vertical displacement:
h = v₀²/2g
Substituting the known values:
h = (30²) / (2 * -9.8)
h = 900 / -19.6
h = -45.92
Again, since height cannot be negative, we find that the rock does not reach a maximum height of 45.92m.
As a result, we cannot determine the velocity of the rock at these heights, or just before it lands.