Question

if you throw a rock straight up at a speed of 19m/s. How long goes it take the ball to reach its maximum height?

Answer 1

Answer:

It requires 1.9 seconds to reach maximum height.

Explanation:

As per given question,  

Initial velocity (U) =19 m/s

Final velocity (V) = 0 m/s

$\text { Taking acceleration due to gravity }(a)=10 \mathrm{m} / \mathrm{s}^{2}$

Maximum height = S

Time taken is "t"

Calculating time taken to reach maximum height:

We know that time taken to reach the maximum height is calculated by using the formula V = U + at

Substitute the given values in the above equation.

Final velocity is “0” as there is no velocity at the maximum height.

$0=19+10 \times t$

$-19=10 \times t$

$\frac{-19}{10}=t$

t = 1.9 seconds.

The time taken to reach maximum height is 1.9 seconds.

Calculating maximum height:

$\text { Consider the equation } V^{2}-U^{2}=2 a S$

Solving the equation we will get the value of S

$0-19^{2}=2 \times(-10) \times \mathrm{S} .(-\text { is due to opposite of gravity) }$

-361 = -20S

Negative sign cancel both the sides.

$\mathrm{S}=\frac{361}{20}$

S = 18.05 m

Maximum height is 18.05 m  .