A stone is thrown horizontally from a tower. In 0.5 second after the stone began to move, the numerical value of its velocity was 5 / 4 times its initial velocity. The initial velocity of stone is v m / s. Determine 3 v

Question

A stone is thrown horizontally from a tower. In 0.5 second after the stone began to move, the numerical value of its velocity was 5 / 4 times its initial velocity. The initial velocity of stone is v m / s. Determine 3 v (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

After 0.5 sec along vertical using v y =u y + a y t v y =0+ g (0.5)=0.5 g Along horizontal, vx=v Using v text net =√ v x 2+ v y 2 √ v 2+0.25 g 2 =1.5 v v 2+0.25 g 2=2.25 v 2 ⇒ 1.25 v 2=(4.9)2 v =4.382 m / s ≈ 4.4 m / s

Q. A stone is thrown horizontally from a tower. In $0.5$ second after the stone began to move, the numerical value of its velocity was $5/4$ times its initial velocity. The initial velocity of stone is $v m / s$ . Determine $3 v$

After $0.5 sec$ along vertical using $v_{y} = u_{y} + a_{y} t$
$v_{y} = 0 + g (0.5) = 0.5 g$
Along horizontal,
$v_{x} = v$
Using $v_{net} = v_{x}^{2} + v_{y}^{2} v^{2} + 0.25 g^{2}$
$= 1.5 v$
$v^{2} + 0.25 g^{2} = 2.25 v^{2}$
$\Rightarrow 1.25 v^{2} = (4.9)^{2}$
$v = 4.382 m / s \approx 4.4 m / s$

Q. A stone is thrown horizontally from a tower. In 0.5 second after the stone began to move, the numerical value of its velocity was 5/4 times its initial velocity. The initial velocity of stone is vm/s. Determine 3v

After 0.5sec along vertical using vy​=uy​+ay​t vy​=0+g(0.5)=0.5g Along horizontal, vx​=v Using vnet ​=vx​2+vy​2​v2+0.25g2​ =1.5v v2+0.25g2=2.25v2 ⇒1.25v2=(4.9)2 v=4.382m/s≈4.4m/s

Q. A stone is thrown horizontally from a tower. In $0.5$ second after the stone began to move, the numerical value of its velocity was $5/4$ times its initial velocity. The initial velocity of stone is $v m / s$ . Determine $3 v$

After $0.5 sec$ along vertical using $v_{y} = u_{y} + a_{y} t$
$v_{y} = 0 + g (0.5) = 0.5 g$
Along horizontal,
$v_{x} = v$
Using $v_{net} = v_{x}^{2} + v_{y}^{2} v^{2} + 0.25 g^{2}$
$= 1.5 v$
$v^{2} + 0.25 g^{2} = 2.25 v^{2}$
$\Rightarrow 1.25 v^{2} = (4.9)^{2}$
$v = 4.382 m / s \approx 4.4 m / s$