A projectile is launched from ground level with an initial velocity of 20 m/s at

Question

A projectile is launched from ground level with an initial velocity of 20 m/s at an angle of 40º above the horizontal (use g = 10 m/s2; choose positive direction up) • The horizontal component of the acceleration the instant after it is launched is 12.96 m/s . • The vertical component of the acceleration the instant after the projectile is launched is 15.32 m/s . • The horizontal component of the acceleration 1 second after it is launched is . • The vertical component of the acceleration 1 second after the projectile is launched is . • The horizontal component of the acceleration at maximum height is . • The vertical component of the acceleration at maximum height is . • The horizontal component of the acceleration 1 second before it lands on the ground is . • The vertical component of the acceleration 1 second before it lands on the ground is . • The horizontal component of the acceleration just before it lands on the ground is . • The vertical component of the acceleration just before it lands on the ground is .

Explanation / Answer

I would assume that you have mistakenly written acceleration in place of velocity at many places. Solution is below

x represent horizontal

y represent vertical

An instant after

Vx= 20*cos40 =15.32 m/s

Vy=12.96 m/s

After 1 sec

Vx= 15.32 m/s (Does not chage as no horizontal force active)

V=U+a*t

Vy=12.96-10*1

=2.96 m.s

At max height

Vx=15.32 m/s  (Does not chage as no horizontal force active)

Vy = 0

Just before it reaches ground

Vx= 15.32 m/s  (Does not chage as no horizontal force active)

Vy = -12.96 m/s (equal to initial velocity but opposite in direction)

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