Projectile Motion Problem Help

  • Thread starter gabbiem10
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  • #1
gabbiem10
2
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I am in an Honors Physics I class at my school and we are just starting to learn about projectile motion. I am having a bit of trouble with some homework and am in need of help.

Problems:
1. I kick a rock horizontally off a ledge that is 20 m high. Once in the air, the rock is in free-fall. How far forward does the rock travel before hitting the ground?
A: I used ΔY=1/2(a)(Δt)^2 to solve for Δt if ΔY = 20m Vy,i = 0 m/s and
a = -9.8 m/s2. I got Δt ≈ 2.0s. Is it even possible to solve this with the given information?

2. A long-jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28 degrees above the horizontal. Determine the long-jumper's time of flight.
I solved for the y component using Vy,i = Rsinθ and
Vx = Rcosθ. Vy,i = 5.6 m/s and Vx = 10.6 m/s. a = -9.8 m/s^2 but that was as far as I could get with the given information.
 

Answers and Replies

  • #2
azizlwl
1,065
10
I am in an Honors Physics I class at my school and we are just starting to learn about projectile motion. I am having a bit of trouble with some homework and am in need of help.

Problems:
1. I kick a rock horizontally off a ledge that is 20 m high. Once in the air, the rock is in free-fall. How far forward does the rock travel before hitting the ground?
A: I used ΔY=1/2(a)(Δt)^2 to solve for Δt if ΔY = 20m Vy,i = 0 m/s and
a = -9.8 m/s2. I got Δt ≈ 2.0s. Is it even possible to solve this with the given information?

2. A long-jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28 degrees above the horizontal. Determine the long-jumper's time of flight.
I solved for the y component using Vy,i = Rsinθ and
Vx = Rcosθ. Vy,i = 5.6 m/s and Vx = 10.6 m/s. a = -9.8 m/s^2 but that was as far as I could get with the given information.

1. Just a nudge, it won't go far forward. Forward depends on initial horizontal velocity.
2. Hidden data. Vertical displacement =0
 
  • #3
gabbiem10
2
0
1. Just a nudge, it won't go far forward. Forward depends on initial horizontal velocity.

The problem is that the initial horizontal velocity is not given neither is the horizontal displacement. The time was calulated base upon other given factors.
 

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