Question

EXPLORE A projectile is launched with a launch angle of 30° with respect to the horizontal direction and with an initial speed of 40 m/s.

(A) How do the vertical and horizontal components of the projectile's velocity vary with time?

(B) How long does it remain in flight?

(C) For a given launch speed, what launch angle produces the longest time of flight?

CONCEPTUALIZE Consider the projectile to be a point mass that starts with an initial velocity, upward and to the right, with forces from air resistance neglected. There is no force acting horizontally to accelerate its horizontal motion, while its vertical motion is accelerated downward by gravity. Therefore, as the projectile moves to the right at a constant rate, the vertical part of its motion consists of first rising upward and then moving downward until the projectile strikes the ground. Use the simulation to display the projectile motion.

CATEGORIZE The velocity has components in both the x- and y-directions, so we categorize this as a problem involving particle motion in two dimensions. The particle also has only a y-component of acceleration, so we categorize it as a particle under constant acceleration in the y-direction and constant velocity in the x-direction.

ANALYZE (A) How do the vertical and horizontal components of the projectile's velocity vary with time? The initial velocity in the x-direction vx0 is related to the initial speed by

vx0 = v0 cos 30°.

The constant velocity in the x-direction means that the equation describing the time dependence of x for the particle, with x0 taken as 0, is

x = x0 + vx0t = 0 + [ input answer ] m/s t).

The equation for the vertical coordinate, which is constantly accelerating downward at g = 9.8 m/s2, is

y = y0 + vy0t − 1 2 gt2 = [ input answer ] m/s t + ( [ input answer ] m/s2) t2.

FINALIZE The − 1 2 gt2 term is negative. The other time-dependent term is proportional to t and positive. Which of the two dominates at small t? Which term's magnitude gets larger faster as t gets large? What effect does that have on the sign of the y-coordinate as t starts out small and then gets larger? Is this consistent with the path you expect the projectile to take?\

ANALYZE (B) How long does it remain in flight? The y-component of the projectile's velocity decreases by 9.8 m/s for each second of flight as the projectile rises. Therefore it takes a time of

ty,max = vy0 g = v0 sin θ / g

for the vertical component of velocity to reach a value of 0, which occurs at the projectile's maximum height. At each height on the way down the particle has regained the same speed and has the same acceleration as it had on the way up, so that the complete time of flight is twice the time to reach the maximum height, and is equal to

tflight = 2v0 sin θ / g.

In the present problem, that expression gives

tflight = [ input answer ] s.

(C) For a given launch speed, what launch angle produces the longest time of flight?

The time of flight for a given initial speed v0,

tflight = 2v0 sin θ / g

is largest when sin θ is largest, which is at θ = [ input answer ] °.

Answer 1

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