Maximum height

Classical Mechanics Level 3

A ball is thrown from a point 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal.Find the maximum height of the ball above the ground.

The answer is 9.5.

2 solutions

Ahmed Hassaneen
May 23, 2014

Height is max , when Vy = 0 , Vy =Voy-gt=0 , t= (Voy)/g = 1.3118 sec
h-ho=(Voy)(t)-0.5gt^2 ==> h =1+ (VoSin(40))(1.3118)- 0.5(9.8)(1.3118)^2= 9.432 meters.

$At\quad the\quad maximum\quad height\quad vertical\quad velocity\quad is\quad 0.\\ Launched\quad from\quad 1m\quad above\quad ground,\\ therefore\quad x=\quad S\quad +\quad 1.\\ Resolving\quad vertically:\\ \\ S\quad =\quad ?\\ U\quad =\quad 20sin40\\ V\quad =\quad 0\\ A\quad =\quad -g\quad (-9.8)\\ T\quad =\quad ?\\ \\ { V }^{ 2 }\quad =\quad { U }^{ 2 }\quad +\quad 2AS\\ 0\quad =\quad ({ 20sin40 })^{ 2 }\quad +\quad 2(-9.8)(S)\\ S\quad =\quad \frac { { (20sin40) }^{ 2 } }{ 19.6 } \quad =\quad 8.432m\\ \\ As\quad x\quad =\quad S+1\quad =\quad 9.432\\$

Jake Sopher - 6 years, 8 months ago

Zineb Moumen
May 24, 2014

Vy(t) = -gt+Voy .. when height is max velocity is null ==> t=(Voy/g)=1, 28 s Y (t) = -(g/2) t^2 + Voy t + h ... ( h=1m) Ym= 9, 43 m

Maximum height

The answer is 9.5.

2 solutions

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