Question

A person standing on a cliff extends their arm past the cliff's edge and throws a stone with velocity 16.0 m/s downward. The stone is 80.0 m above ground when it leaves the person's hand. a) Draw a diagram of the problem and label the known and unknown quantities. b) When will the stone land? c) What is the stone's velocity when it lands? Show your work

Answer 1

Answer:

b) 6.03 seconds

c) 43.164 m/s

Explanation:

t = Time taken

u = Initial velocity = 16 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=16-9.81\times t\\\Rightarrow \frac{-16}{-9.81}=t\\\Rightarrow t=1.63 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=16\times 1.63+\frac{1}{2}\times -9.81\times 1.63^2\\\Rightarrow s=13.05\ m$

So, the stone would travel 13.05 m up

So, total height of the stone would fall is 13.05+80 = 93.05 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 93.05=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{93.05\times 2}{9.81}}\\\Rightarrow t=4.4\ s$

b) The stone will land 1.63+4.4 = 6.03 seconds later

$v=u+at\\\Rightarrow v=0+9.81\times 4.4\\\Rightarrow v=43.164\ m/s$

c) The stone's velocity when it lands is 43.164 m/s