posted by lina on .
A man stands on the roof of a 19.0 m-tall building and throws a rock with a velocity of magnitude 30.0 m/sat an angle of 42.0 degrees above the horizontal. You can ignore air resistance.
Calculate the maximum height above the roof reached by the rock.
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
The vertical component of velocity when the rock is thrown is 30 sin 42 = 20.07 m/s
Calculate the time T that it takes the rock to hit the ground using that initial velocity.
Hint: 20.07 T - (g/2) T^2 + 19.0 = 0
I assume you know what g is, and how to solve that quadratic equation. Take the positive root.
T times the horizontal velocity component is the horizontal distance that the rock travels.
The maximum height h of the rock is attained when the vertical velocity component is zero. This happens when
gt = 20.07 m/s
h = (1/2)(20.07 m/s)^2/g
Ebergy considerations can be used to compute the velocity at impact
Vo = 30 m/s
Xo = 0
Yo = 0
theta = 42
ax = 0 m/s^2 i
ay = -9.8 m/s^2 j
Vox = 30cos42
Vox = ____ m/s i