Answer :
Answer:
Gravitational acceleration of the ball would be [tex](-2)\; {\rm m\cdot s^{-2}}[/tex], assuming that air resistance is negligible, the surface is level, and gravitational attraction is the only force on this ball.
Explanation:
Assume that air resistance is negligible, the surface is level, and that gravitational attraction (downward) is the only force on the ball. While the ball is in flight:
- Horizontal velocity of the ball would be constant.
- Vertical velocity of the ball would not be constant, but the acceleration in the vertical direction would be constantly [tex]a_{y}[/tex].
- Additionally, by the conservation of mechanical energy, the vertical velocity of the ball right before landing ([tex]v_{y}[/tex]) would be the same in magnitude as the initial vertical velocity [tex]u_{y}[/tex], but opposite in direction.
Given the initial velocity [tex]u = 10\; {\rm m\cdot s^{-1}}[/tex] and the angle of elevation of this velocity [tex]\theta = 45^{\circ}[/tex], the initial value of horizontal and vertical velocity can be expressed as:
- Initial horizontal velocity: [tex]u_{x} = u\, \cos\left(\theta\right)[/tex].
- Initial vertical velocity: [tex]u_{y} = u\, \sin\left(\theta\right)[/tex].
In this question, the range [tex]s[/tex] (horizontal distance travelled) of this ball is also given ([tex]s = 50\; {\rm m}[/tex],) and the goal is to find the vertical acceleration [tex]a_{y}[/tex] of this ball during the flight. To do so, apply the following steps:
- Find the duration of the flight [tex]t[/tex] from the horizontal distance travelled ([tex]s[/tex]) and the initial horizontal velocity [tex]u_{x} = u\, \cos\left(\theta\right)[/tex], which stays the same throughout the flight.
- Find the vertical velocity right before landing ([tex]v_{y}[/tex]) using the fact that [tex]v_{y}[/tex] is equal in magnitude to the initial vertical velocity [tex]u_{y}[/tex] but opposite in direction (conservation of mechanical energy.)
- Divide the change in vertical velocity [tex](v_{y} - u_{y})[/tex] by the duration of the flight [tex]t[/tex] to find the vertical acceleration [tex]a_{y}[/tex].
Given that the horizontal distance travelled is [tex]s = 50\; {\rm m}[/tex] and that the horizontal velocity is constantly [tex]u_{x} = u\, \cos\left(\theta\right)[/tex] (same as the initial horizontal velocity,) divide the distance travelled by the speed to find the duration of the motion:
[tex]\displaystyle t = \frac{s}{u_{x}} = \frac{s}{u\, \cos\left(\theta\right)}[/tex].
By the conservation of mechanical energy, the vertical velocity right before landing ([tex]v_{y}[/tex]) would be equal in magnitude to the initial vertical velocity [tex]u_{y}[/tex], but opposite in direction. Therefore:
[tex]v_{y} = -u_{y} = -u\, \sin\left(\theta\right)[/tex].
Vertical acceleration is equal to rate of change in velocity in the vertical direction. Under the assumption that vertical acceleration [tex]a_{y}[/tex] is constant during the flight, divide the change in vertical velocity [tex](v_{y} - u_{y})[/tex] by and the duration of the flight [tex]t[/tex] to find the value of [tex]a_{y}[/tex]:
[tex]\begin{aligned} a_{y} &= \frac{v_{y} - u_{y}}{t} \\ &= \frac{(-u\, \left(\sin(\theta\right)) - u\, \sin\left(\theta\right)}{\displaystyle \frac{s}{u\, \cos\left(\theta\right)}} \\ &= \frac{-2\, u^{2}\, \sin\left(\theta\right)\, \cos\left(\theta\right)}{s} \\&=\frac{-2\, (10\; {\rm m\cdot s^{-1}})^{2}\, \sin\left(45^{\circ}\right)\, \cos\left(45^{\circ}\right)}{50\; {\rm m}}\\&= \frac{-2\, (10\; {\rm m\cdot s^{-1}})^{2}\, \left(\sqrt{2}/2\right)^{2}}{50\; {\rm m}}\\&= (-2)\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
(Negative because the ball is accelerating downwards.)
