At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 52 m horizontally and h = 55 m vertically above the launch point. What are the (a) horizontal and (b)
vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?

Horizontal, v = 52/2 = 26 m/s
Vertical, using s = vt - (1/2)gt^2

v = s/t + (1/2)gt = 37.3 m/s

max height is achieved when vertical velocity is zero

so using v(final) = v (initial) - gt

v(final) = 0 & v(initial) = 37.3

t = v(initial)/g = 3.8 sec

hence D = v(horiz) x t = 26 × 3.8 = 98.8m

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boffeemadrid boffeemadrid · Answer 2 · 2021-09-29T11:08:58+02:00

The horizontal and vertical component of the projectile and the displacement at the maximum height is required.

The horizontal component is 26 m/s the vertical component is 37.31 m/s and the displacement at maximum height is 98.9 m.

[tex]a_y=g[/tex] = Acceleration due to gravity = [tex]-9.81\ \text{m/s}^2[/tex]

d = Horizontal displacement at h = 55 m is 52 m.

h = Vertical height from the ground

[tex]a_x[/tex] = Acceleration in the x axis = 0

t = Time = 2 s

The equations of projectile motion are used here

[tex]d=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 52=u_x\times2+\dfrac{1}{2}\times 0\times 2^2\\\Rightarrow u_x=\dfrac{52}{2}=26\ \text{m/s}[/tex]

The horizontal component is [tex]26\ \text{m/s}[/tex]

[tex]h=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow u_y=\dfrac{h-\dfrac{1}{2}a_yt^2}{t}\\\Rightarrow u_y=\dfrac{55-\dfrac{1}{2}\times(-9.81)\times 2^2}{2}\\\Rightarrow u_y=37.31\ \text{m/s}[/tex]

The vertical component is [tex]37.31\ \text{m/s}[/tex]

[tex]d=\dfrac{u^2\sin2\theta}{2g}\\\Rightarrow d=\dfrac{u^22\sin\theta\cos\theta}{2g}\\\Rightarrow d=\dfrac{u_xu_y}{g}\\\Rightarrow d=\dfrac{26\times 37.31}{9.81}\\\Rightarrow d=98.9\ \text{m}[/tex]

The horizontal displacement at maximum height is [tex]98.9\ \text{m}[/tex].

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