A projectile is launched from ground level to the top of a cliff, which is 195 m away and 155 m high. If the projectile lands on top of the cliff 7.6 seconds after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.
Answer (3)
Refer to the figure shown below.
u = initial launch velocity, m/s θ = launch angle, north of east, deg.
Wind resistance is ignored. g = 9.8 m/s².
The horizontal component of the launch velocity is u cosθ. Because the horizontal distance traveled in 7.6 s is 195 m, therefore 7.6*(u cosθ) = 195 u cosθ = 25.658 (1)
The vertical component of the launch velocity is u sinθ. Because the vertical height traveled in 7.6 s is 155 m, therefore (u sinθ) 7.6 - 0.5 9.8*(7.6²) = 155 u sin θ - 37.24 = 20.395 usin θ = 57.635 (2)
From (1) and (2), obtain tan θ = 57.635/25.658 = 2.246 θ = 66°
From (2), obtain u = 57.635/sin(66°) = 63.09 m/s
Answer: Velocity = 63.09 m/s Direction = 66° north of east (or 66° measured counterclockwise from the x-axis) .
The initial velocity with which the projectile is launched is \fbox{\begin\\63.08\text{ m/s}\end{minispace}} and the projectile is fired at \fbox{\begin\\65.99^\circ\end{minispace}} with repect to ground. ;
The initial velocity of the projectile is approximately 63.08 m/s, and it is launched at an angle of about 65.99° above the horizontal. This was determined using calculations for both horizontal and vertical motion over a time interval of 7.6 seconds. The calculations take into account the horizontal distance to the cliff and the height of the cliff.
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