48. A projectile is shot at a hill, the base of which is 300 m away. The project

Question

48. A projectile is shot at a hill, the base of which is 300 m away. The projectile is shot at 60 above the horizontal with an initial speed of 75 m/s. The hill can be approximated by a plane sloped at 20° to the horizontal. Relative to the coordinate system shown in the following figure, the equation of this straight line is y=(tan20"X-109. Where on the hill does the projectile land? y (tan 20°)x 109 75 m/s 60° 20° 300 m 49. An astronaut on Mars kicks a soccer ball at an angle of 45° with an initial velocity of 15 m/s. If the acceleration of gravity on Mars is 3.7m/s2, (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the same kick on the Moon, where gravity is one-sixth that of Earth?

Explanation / Answer

48.

Break the motion of projectile in to two component ax as 0 and ay as -g .

Take the horizontal distance covered by projectile is x m from the origin,and verticle final position of projectile is y m.

From the equation of the motion

y = yo + vyo*t + 0.5*ay*t2

y = 0+ 75*sin60*t + 0.5*(-9.8)*t2

y = 64.95*t - 4.9*t2

substitute y = (tan20o)x -109 in the above equation

(tan20o)x -109 = 64.95*t - 4.9*t2

(0.36*x)-109 = 64.95*t - 4.9*t2 ..............eq1

now ,we using the following equation

x = xo + vox*t

x = 0 + 75*cos60*t

t = x/75*cos60 = x /37.5

put the value of t in eq1

(0.36*x) - 109 = 64.95*(x/37.5) - 4.9*(x/37.5)2

0.0035x2 - 1.372x - 109 = 0

By solving above equation

x = 459.74 m

so,

y = (tan20o)*459.74 - 109

y = 56.5 m

The projectile hits the plane of the hill at the height of 56.5m from the foot of the hill.

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