Answer:
[tex]57.24^{\circ}[/tex]
24.21 m/s
Explanation:
x = Displacement in x direction = 55 m
y = Displacement in x direction = 0
[tex]y_0[/tex] = Height of the ball when the ball is kicked = 1 m
t = Time taken = 4.2 s
[tex]a_y=g[/tex] = Acceleration due to gravity = [tex]-9.81\ \text{m/s}^2[/tex]
u = Initial velocity of ball
Displacement in x direction is given by
[tex]x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 55=4.2u_x+0\\\Rightarrow u_x=\dfrac{55}{4.2}\\\Rightarrow u\cos\theta=13.1\ \text{m/s}\\\Rightarrow u=\dfrac{13.1}{\cos\theta}[/tex]
Displacement in y direction is given by
[tex]y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 0=1+4.2u\sin\theta+\dfrac{1}{2}\times (-9.81)\times 4.2^2\\\Rightarrow 0=4.2u\sin\theta-85.5242\\\Rightarrow u\sin\theta=20.36\ \text{m/s}\\\Rightarrow u=\dfrac{20.36}{\sin\theta}[/tex]
[tex]\dfrac{13.1}{\cos\theta}=\dfrac{20.36}{\sin\theta}\\\Rightarrow \dfrac{20.36}{13.1}=\dfrac{\sin\theta}{\cos\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{20.36}{13.1}\\\Rightarrow \theta=57.24^{\circ}[/tex]
The ball should be kicked at an angle of [tex]57.24^{\circ}[/tex]
[tex]u=\dfrac{13.1}{\cos\theta}=\dfrac{13.1}{\cos57.24^{\circ}}\\\Rightarrow u=24.21\ \text{m/s}[/tex]
The initial speed of the ball is 24.21 m/s.
The number of unknowns in the question are two therefore two equations
are required to find a solution.
Part A: The angle at which the ball should be kicked is approximately 57.56°
Part B: The initial speed with which the ball should be kicked is approximately 24.41 m/s.
Reasons:
The given parameters are;
Time the ball should be in the air = 4.2 s
Point the ball should land, d = 5.5 m from where it was kicked
Height at which the ball leaves the ground = 1.0 m
Part A;
The angle at which the ball should be kicked.
Solution;
The horizontal distance, d = uₓ·t
Therefore;
[tex]u_x = \dfrac{d}{t} = \dfrac{5.5}{4.2} = \dfrac{55}{42}[/tex]
[tex]u_x = u \cdot cos(\theta)= \dfrac{55}{4.2}[/tex]
Where;
u = The initial velocity
[tex]Vertical \ motion , \ h =\mathbf{h_0 + u_y\cdot t - \dfrac{1}{2} \cdot g \cdot t^2}[/tex]
Which gives;
[tex]0 =1 + u_y\times 4.2 - \dfrac{1}{2} \times 9.81 \times 4.2^2 = 4.2\cdot u_y - 86.5242[/tex]
[tex]u_y =u \cdot sin(\theta) = \dfrac{86.5242}{4.2} = 20.601[/tex]
[tex]\dfrac{u_y}{u_x} =\dfrac{20.601}{\left(\dfrac{55}{4.2} \right)} = \mathbf{ \dfrac{u \cdot sin(\theta)}{u \cdot cos(\theta)}} = tan\left( \theta)[/tex]
[tex]\mathrm{The \ angle \ at \ which \ the \ ball \ should \ kicked } \ \theta = arctan \left(\dfrac{20.601}{\left(\dfrac{55}{4.2} \right)} \right) \approx \underline {57.56 ^{\circ}}[/tex]
Part B
[tex]u \cdot cos(\theta)= \dfrac{55}{4.2}[/tex]
[tex]The \ initial \ speed, \ u= \dfrac{55}{4.2 \times cos(57.56^{\circ })} \approx 24.41[/tex]
The initial speed of the ball, u ≈ 24.41 m/s
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