Two projectiles are launched from a cliff of height H = 16m as shown in the diag

Question

Two projectiles are launched from a cliff of height H = 16m as shown in the diagram below. Projectile 1 is launched with velocity v1 = 12 m/s at an angle q1, where q1 = 35 degrees. Projectile 2 is launched with velocity v2 = 12 m/s at an angle q2, where q2 = 50 degrees. The next six questions pertain to this physical situation. I am just stuck on this entire problem thanks!!

9) If the projectiles are lauched at the same time, which projectile lands first? Projectile 1 lands first. Projectile 2 lands first. The projectiles land at the same time.

10) What is the maximum height of projectile 2 above the ground?(not from the top of the cliff) y = 7.9 m y = 11.3 m y = 20.3 m y = 26.0 m y = 38.4 m

11) At the top of projectile 2's path, what is its speed? v = 0 m/s v = 5.6 m/s v = 7.7 m/s v = 9.8 m/s v = 12.0 m/s

12) How far from the cliff does projectile 2 land from the base of the cliff? x = 19.1 m x = 22.9 m x = 27.5 m x = 30.1 m x = 31.3 m

13) Right before projectile 2 hits the ground, what is its speed? v = 0 m/s v = 12.0 m/s v = 18.7 m/s v = 21.4 m/s v = 30.3 m/s

14) Which projectile lands farther away? Projectile 1 lands farther away. Projectile 2 lands farther away. The projectiles land at the same distance from the cliff.

15) A car's initial velocity is along the -x (negative x) direction. As it slows to a stop, its acceleration: is zero points in the +x direction points in the -x direction

Explanation / Answer

(9) Since the velocity of both projectile is same only the angle is different therefore the projectile with higher angle will
take more time before landing.
Hence projectile 1 will land first.
(10) Max height = (VSin50)2/2g this is from the cliff
   = (12Sin50)2/2g = 4.307 m
From ground = 4.307+16= 20.3 m
(11) At the top of the projectile it will have only horizontal component therefore
velocity will be = VCos50 = 7.713 m/s
(12) First of all we have to calculate the time of the motion
   -16 = VSin50*t - 0.5gt2
4.905t2 - 9.192t - 16 = 0
on solving we get , t = 2.972 seconds
Now the horizontal distance covered = VCos50 *t = 22.9m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.