Question

An object is thrown vertically up and attains an upward velocity of 34 m/s when it reaches one fourth of its maximum height above its launch point. What was the initial speed of the object?

Answer 1

Answer:

39.26 m/s

Explanation:

Let the maximum height at which the object reaches is h.

Let the initial speed f the object is u.

velocity at height h/4 is 34 m/s

acceleration due to gravity = 9.8 m/s^2

For AB part

Use third equation of motion for AB part

here, initial velocity is v = 34 m/s

Final velocity at maximum height, v' = 0

Height = h - h/4 = 3h/4

$v^{2}=u^{2}-2gh$

$0^{2}=34^{2}-2\times 9.8 \times \frac {3h}{4}$

h = 78.64 m

For OA part

initial velocity = u

final velocity, v = 34 m/s

Height = h / 4 = 78.64 / 4 = 19.66 m

Use third equation of motion

$v^{2}=u^{2}-2gh$

$34^{2}=u^{2}-2\times 9.8 \times 19.66$

$u^{2}=1541.336$

u = 39.26 m/s