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2 Motion and forces

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Question 33

A stationary climber accidentally releases a torch. The torch falls for 1.40 s before reaching a ledge below.

Ignore air resistance and use g=10 m/s2g=10\ \mathrm{m/s^2}g=10 m/s2.

Use the equations a=v−ut\displaystyle a=\frac{v-u}{t}a=tv−u​ and v2−u2=2axv^2-u^2=2axv2−u2=2ax.

What is the distance travelled by the torch between 0.60 s and 1.40 s after its release?

[1]
A

8.0 m8.0\ \mathrm{m}8.0 m

B

4.0 m4.0\ \mathrm{m}4.0 m

C

3.2 m3.2\ \mathrm{m}3.2 m

D

9.8 m9.8\ \mathrm{m}9.8 m

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2 Motion and forces Questions

  1. GCSE
  2. /Physics
  3. /2 Motion and forces

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