A small object begins a free-fall from a height of H = 86.5 m at t0 = 0 s. After ? = 2.80 s, a second small object is launched vertically up from the ground with the initial velocity v0 = 41.2 m/s. At what height from the ground will the two objects first meet?

this can be solve using the formulad = vt + 0.5at^2
where d is the distancev is the initial velocitya is the acceleration ( a = - 9.81 m/s2 due to gravity)t is the time
fisrt lets calculate the distance of the small object after 2.80 sd = 0(2.8) + 0.5(9.81)(2.8^2)d = - 38.4552 md = 86.5 - 38.4552 d = 48.0448 m
so at this point let as calculate the distance they are equal and solve the velocity of the small object at t = 2.8vf = v + atvf = 9.81(2.8) = 27.488d = d86.5 - 27.488t - 0.5at^2 = vt + 0.5at^20(t) + 0.5(9.81)t^2 = 41.2t + 0.5(-9.81)t^2solve for tt = 1.26 s
substitute td = = 41.2t + 0.5(-9.81)t^2d = 41.2(1.26) + 0.5(-9.81)(1.26)^2d = 44.11 m

johanrusli johanrusli · Answer 2 · 2019-08-18T14:36:06+02:00

The two objects first meet at about 26.4 m from the ground

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)

v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

Given:

H = 86.5 m

t = 2.8

u = 41.2 m/s

Unknown:

h = ?

Solution:

First Object:

[tex]h_1 = h_o - \frac{1}{2}gt^2[/tex]

[tex]h_1 = H - \frac{1}{2}gt^2[/tex]

[tex]h_1 = 86.5 - \frac{1}{2}(9.81)t^2[/tex]

[tex]\boxed {h_1 = 86.5 - 4.905t^2}[/tex]

Second Object:

[tex]h_2 = ut_2 - \frac{1}{2}gt_2^2[/tex]

[tex]h_2 = u(t - 2.80) - \frac{1}{2}g(t - 2.80)^2[/tex]

[tex]h_2 = 41.2(t - 2.80) - \frac{1}{2}(9.81)(t - 2.80)^2[/tex]

[tex]h_2 = 41.2t - 115.36 - 4.905(t^2 - 5.60t + 7.84)[/tex]

[tex]\boxed {h_2 = -4.905t^2 + 68.668t - 153.8152}[/tex]

Two Objects meet:

[tex]h_1 = h_2[/tex]

[tex]86.5 - 4.905t^2 = -4.905t^2 + 68.668t - 153.8152[/tex]

[tex]68.668t = 86.5 + 153.8152[/tex]

[tex]68.668t = 240.3152[/tex]

[tex]\boxed {t \approx 3.50 ~ s}[/tex]

[tex]h_1 = 86.5 - 4.905t^2[/tex]

[tex]h_1 = 86.5 - 4.905(3.50)^2[/tex]

[tex]\large {\boxed {h_1 \approx 26.4 ~ m} }[/tex]

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration