A) An object is thrown downward with an initial speed of 17 m/s from a height of

Question

A) An object is thrown downward with an initial
speed of 17 m/s from a height of 74 m above
the ground. At the same instant, a second
object is propelled vertically up from ground
level with a speed of 57 m/s .
At what height above the ground will the
two objects pass each other? The accelerationof gravity is 9.8 m/s^2
Answer in units of m?

B) Consider the position of a ball thrown down
with an initial speed of 24 m/s.What will be its position after 3.5 s? Let
the initial position be 0. The acceleration of
gravity is 9.8 m/s^2
Answer in units of m?

Explanation / Answer

A) Let the Y1 and Y2 be the positions of ball1 and ball2.
Considering the acceleration due to gravity as 'g'
We can say that,
Y1 = u1*t + 0.5*g*t^2 (by definition) ----(1)
Y2 = u2*t - 0.5*g*t^2 (by definition)
Where, t is the time any instant ,
u is the initial velocity.
Considering the directions, to find out the positions at which the balls cross,
74-Y1 = Y2
74 – (u1*t + 0.5*g*t^2) = (u2*t - 0.5*g*t^2)
74 – u1*t - 0.5*g*t^2 = (u2*t - 0.5*g*t^2)
74 – u1*t - 0.5*g*t^2 = (u2*t - 0.5*g*t^2)
74 – u1*t = u2*t
t = 74/(u1+u2)
Since u1,u2 are 17 and 57 m/s respectively
Solving the equation gives, t=1 sec.

Substituting ‘t’ in equation (1) gives
Y1 = 21.9m
The balls cross each other at a height of 52.1 m (74 – 21.9).

B) Applying the same definition

Y = u*t + 0.5*g*t^2 (by definition)

Where, t is the time any instant ,
u is the initial velocity.

Y = 24*3.5 +0.5*9.8*3.5^2 = 144.025m.

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