A projectile is launched with an initial speed of 80.0 m/s at an angle of 40.0°

Question

A projectile is launched with an initial speed of 80.0 m/s at an angle of 40.0° above the horizontal. The projectile lands on a hillside 3.60 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(a) What is the projectile's velocity at the highest point of its trajectory?

magnitude    ________  m/s

direction      _________ ° counterclockwise from the +x-axis

(b) What is the straight-line distance from where the projectile was launched to where it hits its target?

______ m

Explanation / Answer

Horizontal componnent of velocity Vox = Vo cos theta

Vox = 80 * cos 40 = 61.28 m/s

so Velocity at its highest point of trajectory is 61.28 m/s i

and its vertical component of Velocity = 0

------------------------------------

b. Xo = Vo t cos theta

y(t) = Vot sin theta - 0.5 gt^2

final position xf = Vo tf cos theta

Distance rom lauch point D = sqrt(xf^2 + yf^2)

D^2 = votf cos theta )^2 + (votfsintheta = 0.5 gtf^2)

D = (80 *3.6 cos 40)^2 + ((80 * 3.6 * sin 40 - (0.5* 9.8* 3.6*3.6))

D = 252 m

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