An object is thrown vertically from a height of 15 m15m at 19 m/s19ms. How long will it take for the object to hit the ground?

1 Answer
Roella W.
Feb 10, 2016

4.55 s4.55s

Explanation:

The vertical distance the object travels is given by the equation
y = v_(y0)t - 1/2g*t^2y=vy0t−12g⋅t2 where v_(y0)vy0 is the vertical component of the initial velocity, gg is the acceleration due to gravity and tt is the time taken to reach the highest point (where the velocity is zero).
The time can be found from the equation v_(yf) = v_(yo) - g*tvyf=vyo−g⋅t by using the fact that v_(yf) = 0vyf=0

Thus 0=19-9.8*t0=19−9.8⋅t
t = 19/9.8 = 1.94 st=199.8=1.94s

The height the object reaches is therefore 19t - 9.8*t^2/219t−9.8⋅t22
=18.4 m=18.4m

We now have to work out how long the object will take to fall 18.4+15 = 33.4 m18.4+15=33.4m

33.4 =0*t + 1/2g*t^233.4=0⋅t+12g⋅t2
t = sqrt(33.4*2/9.8) =2.61 st=√33.4⋅29.8=2.61s

The total time is 1.94 + 2.61 = 4.55 s1.94+2.61=4.55s

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