A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction. (a) What is the projectile’s velocity at the highest point of its trajectory? (b) What is the straight-line distance from where the projectile was launched to where it hits its target?

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be

v_ox=v_o*cosФ

=60*cos (30)

= 51.96 m/s^-1

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onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-16T22:51:12+01:00

(a) The projectiles velocity at the highest point of its trajectory is 51.96 m/s.

(b) The maximum height reached by the projectile is 46 m.

The given parameters;

initial speed of the projectile, u = 60 m/s
angle of inclination of the projectile, θ = 30⁰
time of motion of the projectile, t = 4 s

The projectiles velocity at the highest point of its trajectory is calculated as;

[tex]v_f_y = 0\\\\v_f_x = v_i_x = vcos(\theta) = 60 \times cos(60) = 51.96 \ m/s\\\\v_f = \sqrt{v_f_y^2 + v_f_x^2} \\\\v_f = \sqrt{(0)^2 \ + \ (51.96)^2} \\\\v_f = 51.96 \ m/s[/tex]

The maximum height reached by the projectile is calculated as follows;

[tex]H = \frac{u^2 sin^2 \theta }{2g} \\\\H = \frac{ (60^2) \times (sin\ 30)^2}{2\times 9.8} \\\\H = 46 \ m[/tex]

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