A projectile is launched with a launch angle of 65° with respect to the horizont

Question

A projectile is launched with a launch angle of 65° with respect to the horizontal direction and with initial speed 22 m/s.

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

The initial velocity in the x-direction vxi is related to the initial speed by

vxi = vi cos 65°

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 + vxit = 0 + ____ m/s t

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

y = y0 + vyit - 1/2gt2 = ( ____ m/s) t + ( ____ m/s2) t2

(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

ty1max = vy1/g = vi sin /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 = 2vi sin /g

In the present problem, that expression gives tflight = ____ 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 vi,

tflight = 2vi sin /g

is largest when sin is largest, which is at = ____ °

Explanation / Answer

a) vx = ux = ucos65 = 9.29 m/s

vy = usin65-gt

total time t = 2usin65/g = 4.07 s

b) t = 2usin65/g = 4.07 s

c) if theta becomes 90 only sin angle is more

so theta = 90

t = 2u/g = 2*22/9.8 = 4.49 sec

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