Answer:
The initial speed of the projectile is 52.78 m/s
Explanation:
Applying equations of motion
The horizontal distance traveled by the projectile = R
R = u²sin2θ/g
where;
u is the initial speed of the projectile
g is acceleration due to gravity = 9.8
θ is the angle of projection = 19.0°
and given R = 175m
175 = u²sin2θ/g
175 = u²sin(2*19)/9.8
u²sin38⁰ = 1715
u² * 0.61566 = 1715
u² = 1715/0.61566
u² = 2785.628
u = √2785.628
u = 52.78 m/s
The initial speed of the projectile is 52.78 m/s
58.9m/s
A projectile motion is a two-way motion - horizontal and vertical.
Using the equation of motion;
h = ut + [tex]\frac{1}{2}[/tex] x a x t² -----------------(i)
Where;
h = vertical/horizontal distance of the object in motion
u = initial velocity of the object
a = acceleration of the body
t = time taken in motion.
The equation can further be resolved into its horizontal and vertical components as follows;
=>Horizontal component of the motion
s = (ucosθ x t) + ([tex]\frac{1}{2}[/tex] x a x t²) ----------------(ii)
where the vector quantity u from equation(i), has been replaced with its horizontal component ucosθ and h has been replaced with s to denote the horizontal distance.
Remember that for a projectile motion, the acceleration a, of the horizontal motion is zero(0). Therefore, equation(ii) becomes;
s = ucosθ x t --------------------(iii)
Where s is the horizontal distance covered by the object.
=>Vertical component of the motion
h = (usinθ x t) + ([tex]\frac{1}{2}[/tex] x a x t²) ------------------(iv)
where the vector quantity u from equation(i), has been replaced with its vertical component usinθ.
Remember that for a projectile motion, the acceleration a, of the vertical motion is constant and is given by the acceleration due to gravity, g. Therefore, equation(iv) becomes;
h = (usinθ x t) + ([tex]\frac{1}{2}[/tex] x g x t²) -------------------------(v)
where g will be negative as the projectile object moves upward against gravity and h is the vertical distance of the object.
Therefore, the equation (v) becomes;
h = (usinθ x t) - ([tex]\frac{1}{2}[/tex] x g x t²) -----------------------------(vi)
Given from question;
θ = angle of projection = 19.0°
h = vertical distance = 11.0m
s = horizontal distance = 175m
Substitute these values into equations (iii) and (vi) as follows;
s = ucosθ x t
=> 175 = u x cos19° x t
=> ut x cos19.0° = 175
=> ut x 0.946 = 175
=> ut = 175 / 0.946
=> ut = 184.99 -----------------------------(vii)
Also,
h = (usinθ x t) - ([tex]\frac{1}{2}[/tex] x g x t²) --------- take g = 10m/s²
=> 11 = u sin 19.0° x t - ([tex]\frac{1}{2}[/tex] x 10 x t²)
=> 11 = u x 0.326 x t - (5 x t²)
=> 11 = ut x 0.326 - 5t² ---------------------(viii)
Substitute ut = 184.99 from equation(vii) into equation (viii)
=> 11 = 184.99 x 0.326 - 5t²
=> 11 = 60.31 - 5t²
=> 5t² = 60.31 - 11
=> 5t² = 49.31
=> t² = 49.31 / 5
=> t² = 9.862
=> t = [tex]\sqrt{9.862}[/tex]
=> t = 3.14s
Substituting t = 3.14s into equation (vii) gives;
ut = 184.99
u(3.14) = 184.99
u = 184.99 / 3.14
u = 58.9m/s
Therefore, the initial speed of the projectile is 58.9m/s
