Question

A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the distance (in meters) of the stone above ground level at time t. h(t) = −4.9t2+750 (b) how long does it take the stone to reach the ground? (round your answer to two decimal places.) 12.37 s (c) with what velocity does it strike the ground? (round your answer to one decimal place.) -121.2 m/s (d) if the stone is thrown downward with a speed of 6 m/s, how long does it take to reach the ground? (round your answer to two decimal places.)

Answer 1

(a) The stone moves by uniform accelerated motion, with constant acceleration $g=9.81 m/s^2$ directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
$y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2$

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
$750-4.9 t^2 = 0$
from which we find the time t after which the stone reaches the ground:
$t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s$

(c) The velocity of the stone at time t can be written as
$v(t) = -gt$
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
$v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s$

(d) if the stone has an initial velocity of $v_0 = 6 m/s$, then its law of motion would be
$y(t)=y_0 - v_0t - \frac{1}{2}gt^2$
and we can find the time it needs to reach the ground by requiring again y(t)=0:
$0=750 - 6t - 4.9 t^2$
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.