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During WWI, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 1700 m/s, and was shot with an inclination of 55 degrees. In order to hit the target, adjustments were made for air resistance. Ignoring frictional effects, calculate the following:

a.The Horizontal and Vertical Velocities

b.How far away did the shell hit?

c. How long was the shell in the air?

User Tenesha
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Answer:


975.1\ \text{m/s} and
1392.56\ \text{m/s}


276830.96\ \text{m}

284 seconds

Step-by-step explanation:

(a)
u=\text{Initial velocity}=1700\ \text{m/s}


\theta = Angle of inclination =
55^(\circ)

Horizontal velocity is given by


u_x=u\cos\theta\\\Rightarrow u_x=1700* \cos55^(\circ)\\\Rightarrow u_x=975.1\ \text{m/s}

Vertical velocity is given by


u_y=u\sin\theta\\\Rightarrow u_x=1700* \sin55^(\circ)\\\Rightarrow u_x=1392.56\ \text{m/s}

The horizontal and vertical velocities are
975.1\ \text{m/s} and
1392.56\ \text{m/s} respectively.

(b) Range of projectile is given by


R=(u^2\sin2\theta)/(g)\\\Rightarrow R=(1700^2\sin(2* 55^(\circ)))/(9.81)\\\Rightarrow R=276830.96\ \text{m}

The shell hit the ground
276830.96\ \text{m} from the launch position.

(c) Time of flight is given by


t=(2u\sin\theta)/(g)\\\Rightarrow t=(2* 1700* \sin55^(\circ))/(9.81)\\\Rightarrow t=284\ \text{s}

The shell was in the air for 284 seconds.

User ThomasR
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