Question

How fast would you have to launch a ball at 45 degrees above the horizontal to reach a height of 49 meters in the air? [Must show work]

Answer 1

Let say the ball is projected in air with speed "v" at an angle of 45 degree

now the two components of its velocity will be given as

$v_x = vcos45 = 0.707 v$

$v_y = vsin45 = 0.707 v$

now the maximum height reached by the ball is 49 m

so as it will reach to maximum height its velocity in y direction will become zero

so we can use kinematics in y direction

$v_f^2 - v_i^2 = 2 a y$

$0 - (0.707v)^2 = 2(-9.8)(49)$

$0.5v^2 = 960.4$

$v = 43.8 m/s$

so the speed with which ball is projected upwards must be 43.8 m/s