A projectile is launched with an initial speed of 45.0 m/s at an angle of 33.0°

Question

A projectile is launched with an initial speed of 45.0 m/s at an angle of 33.0° above the horizontal. The projectile lands on a hillside 3.65 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?

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

Explanation / Answer

a) u = 45 m/s

theta = 33 degree

t = 3.65 s

a) at the highest point

velocity of the projectile = u * cos(theta)

velocity of the projectile = 45 * cos(33 degree)

velocity of the projectile = 37.7 m/s

the direction of velocity is horizontal

hence , 0 degree with x axis

b)
at t = 3.65 s

y = 45 * sin(33 degree) * 3.65 - .0.50 * 9.8 * 3.65^2

y = 24.2 m

x = u * cos(theta) * t

x = 45 * cos(33) * 3.65

x = 137.7 m

straight line distance = sqrt(x^2 + y^2)

straight line distance = 139.9 m

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