Question

An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at the same instant. What was the initial speed of the second object? An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at the same instant. What was the initial speed of the second object? 19.6 m/s 21.3 m/s 14.6 m/s 9.90 m/s 4.91 m/s

Answer 1

Answer:

$v_{o}=-14.60m/s$

Explanation:

Kinematics equation for first Object:

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

but:

$v_{o}=0m/s$       The initial velocity is zero

$y_{o}=20m$

it reach the water at in instant, t1, y(t)=0:

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

$t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s$

Kinematics equation for the second Object:

The initial velocity is zero

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

but:

$y_{o}=20m$

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

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

$v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s$

The velocity is negative, because the object is thrown downwards.

Answer 2

Based on the data provided, the initial velocity of the second object is 14.6 m/s.

What was the initial speed of the second object?

Using the equation of motion below:

• h = ut + 1/2gt²

where h is height

• u is initial velocity
• t is time
• g is acceleration due to gravity = 9.81 m/s²

For the first object:

• u = 0
• h = 20 m
• t = ?
• g = 9.81 m/s²

solving for t;

t = √(2h/g)

t = √(2 * 20/9.81)

t = 2.02 s

For the second object:

• t = 2.02 - 1.00 = 1.02 s
• h = 20 m
• u = ?
• g = 9.81 m/s²

u = (h - 1/2gt²)/t

u = (20 - 0.5 * 9.81 * 1.02²)/1.02

u = 14.6 m/s

Therefore, the initial velocity of the second object is 14.6 m/s.

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