Question

With what speed must a ball be thrown vertically up in order to rise to a maximum height of 45 m? And

for how long will it be in air? ​

Answer 1

${v}^{2} = u {}^{2} + 2as$

at top most point v=0

0=u2 +2 ×10×45

${u} = \sqrt{2 \times 10 \times 45}$

u=30 m/s.

for time taken

v=u +at

0=30 + 10t

t=3s

Answer 2

$\huge{\bold{Answer}}$

Given=>

▶ H = 45m

▶g = 9.8$ms^{-2}$

since ball is thrown vertically up, angle will be 90°

For maximum height..

H = $u^{2}\times sin^{2}tita\frac{1}{2g}$

substituting ...

45 = $u^{2}\times sin^{2}90\frac{1}{2\times 9.8}$

45 = $u^{2}\frac{1}{2\times 9.8}$

45 = $\frac{u^{2}}{2\times 9.8}$

$882$ = $u^{2}$

882 = $u^{2}$

u = √882

u = 29.69$m\times s^{-1}$

Time of flight..

but time...

T = $2usintita\frac{1}{g}$

T = $2\times 29.69\times 1\frac{1}{9.86}$

T = 6.06 seconds...

####be Brainly..