Question
A person standing on the top of a cliff, 171 ft high, has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall? Give g = 32 ft/s2.
Answer: Option D
:
D
In the given figure we can find θ by using trigonometry.
tanθ = 171228 = 34
⇒ θ = 37∘.
This means the ball was thrown at an angle 37∘ to horizontal also its given that it was thrown with velocity 15 ft/s.
The dotted line shows how the trajectory of the ball thrown
So we know 2 D is nothing but 2 one dimensional motion.
Let's break the components:
uy = −15sin37∘ = −15×35 = −9ft/s
ay = −32ft/s2
sy = −171ft.
⇒ sy = uyt+12ayt2
−171 = −9t−322t2
16t2+9t−171=0
16t2+57t−48t−171=0
⇒ t = 3sec
We need to know in that time the projectile covered how much of horizontal distance.
So ux=15cos37∘ = 15×45 = 12ft/s
ax = 0; t = 3
sx = uxt = 12×3 = 36ft
⇒ The packet went only 36 ft towards his friend
⇒ The packet fell short by (228−36)
= 192ft.
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:
D
In the given figure we can find θ by using trigonometry.
tanθ = 171228 = 34
⇒ θ = 37∘.
This means the ball was thrown at an angle 37∘ to horizontal also its given that it was thrown with velocity 15 ft/s.
The dotted line shows how the trajectory of the ball thrown
So we know 2 D is nothing but 2 one dimensional motion.
Let's break the components:
uy = −15sin37∘ = −15×35 = −9ft/s
ay = −32ft/s2
sy = −171ft.
⇒ sy = uyt+12ayt2
−171 = −9t−322t2
16t2+9t−171=0
16t2+57t−48t−171=0
⇒ t = 3sec
We need to know in that time the projectile covered how much of horizontal distance.
So ux=15cos37∘ = 15×45 = 12ft/s
ax = 0; t = 3
sx = uxt = 12×3 = 36ft
⇒ The packet went only 36 ft towards his friend
⇒ The packet fell short by (228−36)
= 192ft.
Was this answer helpful ?
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