Question

You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 24.5° from the ground with an initial speed of 26.5 m/s. At what height above its initial position does the softball hit the wall? In your calculations, neglect any effects of air resistance. Number 4.42 m

Answer 1

Solution-

let us assume at time t it hits the wall

=> 26.5*cos(24.5)*t=17.5

=> t= 0.725 seconds

Now the height at which it hits the wall:

26.5*sin(24.5)*t-0.5*9.8*t^2

=> 26.5*sin(24.5)*0.725-0.5*9.8*(0.725)^2

=5.39 m