Question

a ball is thrown upward with a velocity of 100m/s find the maximum height reached by the ball and the time taken to come back to the thrower .​

Answer 1

$\huge\red{\boxed{Hello!!!}}$

$Given\: initial\:velocity = 100ms^{-1}</p><p>$

$Acceleration \:due \:to \:gravity = 10ms^{-2}$

$\bold{\boxed{Formula:h_{max} = \dfrac{u^{2}}{2g}}}$

$h_{max} = \dfrac{100^{2}}{2× 10} = 500m$

$so,\:the \:height \:of \:the \:tower \:is \:500m$

$\bold{\boxed{Formula:h = \dfrac{1}{2}gt^{2}}}$

$Given\: height \:of \:the \:tower = 500m$

$Acceleration \: due\: to\: gravity = 10ms^{-2}$

$\bold{\boxed{500 = \dfrac{1}{2}×10t^{2} = 10sec}}$

$Thus,\:the \:time \:taken\: to\: reach\: the\: floor \:is\: 10sec$

Answer 2

Answer:

Maximum height = 500 metres

Time = 4.2 seconds

Explanation:

Given:

• Initial velocity of the ball = 100 m/s
• Acceleration due to gravity = 10 m/s²

To find:

• Maximum height reached by the ball
• Time taken for the ball to come back to the initial position

Maximum height = u²/2g

Maximum height = 100²/20

Maximum height = 10000/20

Maximum height = 500 metres

Now using second equation of motion:

S=ut +1/2 at²

500=100t+1/2 × 10 × t²

5t²+100t-500=0

By solving the above equation,

t = 4.2 seconds (approx)