Question

What is the distance a ball falls after 1,2,3,4,5 seconds (assumeg is rounded to 10
m/?)?
A) 10, 20, 30, 40, 50 m
B) 5, 20, 45, 80,125 m
C) 5, 10,15, 20, 25 m
D) 10, 25, 60, 90,120 m

Answer 1

Answer:

Let’s review the 4 fundamental kinematic equations of motion for constant acceleration (strongly recommend you commit these to memory – they will serve you well):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, we know u = 0 (assumed), g = 10m/s^2 (given) and we want to know s at several values of t, so we use equation (1):

s = ut + ½at^2s1 = 0 + 5(1^1) = 5m

s2 = 0 + 5(2^2) = 20m

s3 = 0 + 5(3^2) = 45m

s4 = 0 + 5(4^2) = 80m

s5 = 0 + 5(5^2) = 125m

Please mark me as brainliest

Thank you and have a nice day

Answer 2

Answer:

The distance a ball falls after 1, 2, 3, 4, 5 seconds is B) 5, 20, 45, 80,125 m.

Explanation:

Given:

t₁ = 1s, t₂ = 2s, t₃ = 3s, t₄ = 4s, t₅ = 5s

a = g = 10m/s²

To find:

Distances s₁, s₂, s₃, s₄, and s₅

Formula:

s = ut + (1/2)at²

Solution:

From the equations of motion, we have

s = ut + (1/2)at²

Assuming that u = 0, we get

s = 0 + (1/2)at²

s = (1/2)at²

For t₁ = 1s,

s = (1/2)(10)(1²)

s = (1/2)(10)

s = 5m

For t₂ = 2s,

s = (1/2)(10)(2²)

s = (5)(4)

s = 20m

For t₃ = 3s,

s = (1/2)(10)(3²)

s = (5)(9)

s = 45m

For t₄ = 4s,

s = (1/2)(10)(4²)

s = (5)(16)

s = 80m

For t₅ = 5s,

s = (1/2)(10)(5²)

s = (5)(25)

s = 125m

Therefore, the distances are 5m, 20m, 45m, 80m, and 125m.

#SPJ2