A man throws a ball vertically upward and it rises through 40 m and returns to his hands, what was the ascent initial velocity of the ball and for how much time (T) it remained in the air?

Question

A man throws a ball vertically upward and it rises through 40 m and returns to his hands, what was the ascent initial velocity of the ball and for how much time (T) it remained in the air? (A) T=5 sec, u=20 √3 m/s (B) T=5.64 sec, u=10 √2 m / s (C) T=5.64 sec, u=20 √2 m / s (D) None of these

Tardigrade Admin · Accepted Answer

Velocity of descent or ascent always equal.

V = 2 \times g \times h = 2 \times 10 \times 40 = 800 = 202 m / s

Time of ascent and descent always equal.

T =

time of ascent

+

time of descent

T = t_{1} + t_{2}

Let

t

, is the time of ascent and

t,

be that of descent

v = u - g t

0 = u - g t_{1} v = u + g t_{2}

u = g t_{1} u = 0 + g t_{2}

t_{1} = \frac{u}{g} \frac{u}{g} = t_{2}

T = t_{1} + t_{2}

T = \frac{u}{g} + \frac{u}{g}

= \frac{2 u}{g} = \frac{2 \times 800}{10}

T = 4 \times 2

sec

T = 4 \times 1.41 = 5.64

seconds

Q. A man throws a ball vertically upward and it rises through $40 m$ and returns to his hands, what was the ascent initial velocity of the ball and for how much time $(T)$ it remained in the air?

$T = 5$ sec, $u = 203$ m/s

$T = 5.64$ sec, $u = 102 m / s$

$T = 5.64$ sec, $u = 202 m / s$

None of these

Q. A man throws a ball vertically upward and it rises through 40m and returns to his hands, what was the ascent initial velocity of the ball and for how much time (T) it remained in the air?

T=5 sec, u=203​ m/s

T=5.64 sec, u=102​m/s

T=5.64 sec, u=202​m/s

None of these

Q. A man throws a ball vertically upward and it rises through $40 m$ and returns to his hands, what was the ascent initial velocity of the ball and for how much time $(T)$ it remained in the air?

$T = 5$ sec, $u = 203$ m/s

$T = 5.64$ sec, $u = 102 m / s$

$T = 5.64$ sec, $u = 202 m / s$