Question

An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle of 23 degrees BELOW the horizontal. How long, in seconds, does it take to reach the ground

Answer 1

Answer:

$3.58\:\mathrm{s}$

Explanation:

We can use the kinematics equation $\Delta y=v_it+\frac{1}{2}at^2$ to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

$\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}$

Now we can substitute values in our kinematics equation:

• $\Delta y=-100$
• $a=-9.8\:\mathrm{m/s^2}$ (acceleration due to gravity)
• $v_i=-10.3283743813\:\mathrm{m/s}$
• Solving for $t$

$-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}$