Question

A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistance. At the location of the bridge g has been measured to be 9.83 m/s2. (a) If you drop the stone, how long does it take for it to fall to the base of the gorge? (b) If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground? (c) If you throw the stone with a velocity of 20.0 m/s at 30.0° above the horizontal, how far from the point directly below the bridge will it hit the level ground?

Answer 1

Answer:

a)$t_{1}=3.49s$

b)$t_{2}=2.00s$

c)Xmax=80.71m

Explanation:

a)Kinematics equation for the Stone, dropped:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$       initial position is bridge height

$v_{o}=0m/s$       the stone is dropped

The ball reaches the ground, y=0, at t=t1:

$0=h-1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2h/g}=\sqrt{2*60/9.83}=3.49s$

b)Kinematics equation for the Stone, with a initial speed of 20m/s:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$       initial position is bridge height

$v_{o}=-20m/s$       the stone is thrown straight down

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$0=60-20t_{2}-1/2*9.83*t_{2}^{2}$

t2=-6.01       this solution does not have physical sense

t2=2.00

c)Kinematics equation for the Stone, with a initial speed of 20m/s with an angle of 30° above the horizontal:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$       initial position is bridge height

$v_{o}=20sin(30)=10m/s$       the stone is thrown with an angle of 30° above the horizontal

The ball reaches the ground, y=0, at t=t3:

$0=h+v_{o}t_{3}-1/2*g*t_{3}^{2}$

$0=60+10t_{3}-1/2*9.83*t_{3}^{2}$

t3=-2.62       this solution does not have physical sense

t3=4.66

the movement in x:

v=constant=20cos(30)m/s

x(t)=v*t

Xmax=v*t3=20cos(30)*4.66=80.71m