A ball is fired at a speed of 8 m/s at 20˚ above the horizontal from the top of a 5 m high wall. How far from the base of the ball will the ball hit the ground? Also calculate the angle below horizontal the velocity vector makes.
time in air:
hf=hi+vi*t-1/2 (9.8)t^2
0=5+8sin20*t-4.9t^2
solve that quadratic, use the quadratic equation.
Now, distance from base:
d=8Cos20*timeinair.
the angle: at impact?
arctan(verticalveloicty/horizontal)
verticalvelocity..
vf=8sin20-9.8t^2
so
angle arctan(vf/8cos20)
To find the horizontal distance the ball will travel before hitting the ground, we can break down the motion of the ball into its horizontal and vertical components.
1. Vertical Component:
The initial vertical velocity is given by the formula: V_0y = V_0 * sin(θ),
where V_0 is the initial velocity of the ball (8 m/s) and θ is the angle above the horizontal (20˚).
So, V_0y = 8 m/s * sin(20˚) = 2.75 m/s.
We can use this vertical velocity to find the total time of flight using the second equation of motion: Δy = V_0y * t + (1/2) * g * t^2,
where Δy is the vertical displacement (5 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Rearranging the equation gives: 0 = -5 m + (2.75 m/s) * t - (4.9 m/s^2) * t^2,
which is a quadratic equation. Solving this equation will give us the time when the ball hits the ground.
2. Horizontal Component:
The horizontal velocity remains constant throughout the motion. It is given by V_0x = V_0 * cos(θ),
where V_0x is the initial horizontal velocity.
So, V_0x = 8 m/s * cos(20˚) = 7.55 m/s.
Once we have the time of flight from step 1, we can multiply it by the horizontal velocity to get the horizontal distance traveled: Δx = V_0x * t.
Finally, to find the angle below the horizontal the velocity vector makes, we can use the equation: tan(φ) = V_0y / V_0x,
where φ is the angle below the horizontal.
Solving this equation will give us the angle below the horizontal the velocity vector makes.
Following these steps, you can calculate the values.