I\'m not sure how to find magnitude and direction in this question, please show

Question

I'm not sure how to find magnitude and direction in this question, please show work and how you got. The only parts i need help is (c), will rate answer

A projectile is fired with an initial speed of 33 m/s at an angle of 67 above the horizontal. The object hits the ground 8.1 s later. (Neglect air resistance.) (a) How much higher or lower is the launch point relative to the point where the projectile hits the ground? (If the point where the projectile hits the ground is lower, enter a negative number.) 75.43m (b) To what maximum height above the launch point does the projectile rise? 47.1m (c) What are the magnitude and direction of the projectile's velocity at the instant it hits the ground? magnitude 64 direction m/s (counterclockwise from the +x-axis)

Explanation / Answer

given initial speed of projectile, v = 33 m/s

angle, thetsa = 67 deg

time of flight T = 8.1 s

c. from part a, the point of firing is 75.43 m above the point where the projectile hits the ground

so, at maximum height, let the height from the point of contact of projectile to the ground ultimately be h

then

2*g*(h - 75.43) = (vsin(theta))^2

h = 122.46 m

so vertical speed at final location = vy

2*g*h = vy^2

vy = 49.017 m/s

horizontal spoeed, vx = vcos(theta) = 12.894 m/s

final speed = sqroot(vx^2 + vy^2) = 50.684 m/s

angle with horizontal = phi

tan(phi) = -vy/vx

phi = -75.26 deg CCW from +x axis

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