contestada

A rocket is fired at a speed of 92.0 m/s from ground level, at an angle of 48.0 ° above the horizontal. The rocket is fired toward an 15.0-m high wall, which is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

Respuesta :

Answer:14.03 m

Explanation:

Given

initial velocity(u)=92 m/s

Launch angle[tex]=48^{\circ}[/tex]

Height of wall=15 m

Located at x=27 m

maximum height [tex]h=\frac{u^2sin^\theta }{2g}=\frac{92^2(sin48)^2}{2\times 9.8}[/tex]

[tex]h_{max}=238.48 [/tex]

We know equation of trajectory of projectile is

[tex]Y=x\tan\theta -\frac{gx^2}{2u^2(cos\theta )^2}[/tex]

for x=27 m

[tex]y=27\tan48-\frac{9.8\times 27^2}{2\times 92^2\times (cos48)^2}[/tex]

y=29.98-0.942=29.03 m

Thus it passes at a height of 29.03-15 =14.03 m above wall

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