Question

The acceleration due to gravity on Earth is 32 ft/sec2. A tomato is dropped from 100 feet above the ground. Give units in your answers. (a) At what speed does the tomato hit the ground? (b) How long does it take for the tomato to travel the last 10 feet? Give your answer as a decimal approximation with units.

Answer 1

Answer:

a) the final velocity is v=80 ft/s

b) the time required is 2.5 seconds

Step-by-step explanation:

The final velocity is related with the acceleration through

v² = v₀² + 2*g*h

where

v₀= initial velocity= 0 (dropped from test)

g = acceleration

h = height

replacing values

v² = v₀² + 2*g*h = 0² + 2*32 ft/s² * 100 ft = 6400  ft²/s²

then

v= √6400  ft²/s² = 80 ft/s

the tomato hit the ground at v=80 ft/s

h₀= h + v₀*t - 1/2*g*t²  

since h₀=0 ( initial height) and v₀=0

0= h - 1/2*g*t²  

t = √(2*h/g)

replacing values

t = √(2*h/g) = √(2*100 ft / 32 ft/s²) = 10/4 s =2.5 s

t =2.5 seconds

Answer 2

It takes 0.1 s for the tomato to travel the last 10 feet.

Using the equations of motion under gravity, we have the following information;

• Acceleration due to gravity on Earth = 32 ft/sec2

• Height = 100 feet above the ground

From;

v ^2= u^2 + gh

Recall that the tomato was dropped from a height above the ground so u=0 feet/s

v ^2=  gh

v ^2= 32 ft/sec2 × 100 feet

v = √32 ft/sec2 × 100 feet

v = 56.6 ft/s

The time taken to hit the ground is obtained from;

h = ut + 1/2gt^2

but u =0 ft/s

h =  1/2gt^2

Where h = 100 feet

100 feet = 1/2 ×  32 ft/sec2 × t^2

t^2 = 100 feet × 2/32 ft/sec2

t = √100 feet × 2/32 ft/sec2

t = 2.5 s

To travel 90 feet

h = ut + 1/2gt^2

but u =0 ft/s

h =  1/2gt^2

Where h = 90 feet

90 feet = 1/2 ×  32 ft/sec2 × t^2

t^2 = 90 feet × 2/32 ft/sec2

t = √90 feet × 2/32 ft/sec2

t = 2.4 s

Time taken to travel the last 10 feet = 2.5 s - 2.4 s

= 0.1 s

Hence, it takes 0.1 s for the tomato to travel the last 10 feet.

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