Refer to the diagram shown below.
g = 9.8 m/s², and air resistance is ignored.
Let us consider the two trajectories.
Case A; The ball is launched 40° below the horizontal
u = (13 m/s)*cos 40° = 9.9586 m/s, the horizontal velocity
v = -(13 m/s)*sin 40° = -8.3562 m/s
When the ball is at 5 m above ground, it would have traveled 10 m.
The vertical downward velocity is
V² = (-8.3562 m/s²)² + 2*(-9.8 m/s²)*(-10 m) = 265.826
V = +/-16.304 m/s => V = -16.304 m/s (downward)
The horizontal velocity remains unchanged.
The speed of the ball is
√[(-16.304)² + 9.9586²] = 19.105 m/s
Case B: The ball is launched 40° above the horizontal.
u = 9.9586 m/s, as in case A
v = 8.3562 m/s
When the ball is 5 m above ground, the vertical velocity, V, is given by
V² = (8.3562 m/s)² + 2*(-9.8 m/s²)*(-10 m) = 265.826
V = +/- 16.304 m/s
The speed of the ball is
√[(-16.304)² + 9.9586²] = 19.105 m/s
Surprisingly, both speeds are the same.
Answer:
The speed of the ball when it is 5 m above ground is 19.1 m/s (both cases).