When a Springbok is startled it will leap straight up into the air. It extends its legs forcefully, accelerating at 35 m/s^2 through 0.7 m as its legs straighten. Legs fully extended, it leaves the ground and rises straight into the air. With this information:
a. determine with what speed the Springbok leaves the ground
b. determine how far up in the air the Springbok reaches (consider upwards to be positive)
Oh also, consider the moment the Springbok leaves the ground to be t=0 s.
My physics teacher made this so confusing and she said that during the 0.7 meters he is still on the ground but we have to use that number in one of the kinematics equations but I have no clue what the 0.7 would represent.
She also said to use g, which is 9.8 m, but if g and a is used interchangeably in the position equation which one do I use? or in the time-independent equation? I don't know :(
a. determine with what speed the Springbok leaves the ground
b. determine how far up in the air the Springbok reaches (consider upwards to be positive)
Oh also, consider the moment the Springbok leaves the ground to be t=0 s.
My physics teacher made this so confusing and she said that during the 0.7 meters he is still on the ground but we have to use that number in one of the kinematics equations but I have no clue what the 0.7 would represent.
She also said to use g, which is 9.8 m, but if g and a is used interchangeably in the position equation which one do I use? or in the time-independent equation? I don't know :(
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There are a couple of motion equations that will solve this and many other problems, provided acceleration is constant.
1) s = distance = 0.5 a t^2
2) v = a * t
3) average v = v at end minus v at start, divided by 2.
The 35m/s^2 and the 0.7 m leap give enough information to solve the problem. First, you want to know what velocity the bok is going when its feet leave the ground.
The givens are acceleration and distance. Equation 1 uses those terms. But it needs to solve for t. If you do a little algebra, you will find t = sqrt(2s/a). Putting in our values, you get sqrt(2*0.7/35) = 0.2 sec. Thinking about it, that sounds just about right.
Why do I use 35m.s^2 as upward acceleration, but ignore gravity? Because the problem states it is accelerating upward at that rate. Gravity could be anything or the bok could be out in space under no gravity, but it would still spring at that acceleration and its feet would leave the surface in 0.2 sec. In space, it would have to press less hard to accelerate at 35m/s^2 than it would off the earth.
Now the next phase is free flight. You want to know the velocity. Use equation 3 (works when a =constant): V = a*t = 35m/s^2 * 0.2s = 7 m/s.
a) 7 m/s is your answer for your a).
Ok, it is traveling up at 7 m/s, and starts its free flight 0.7 meters off the ground. Use equation 2 to find out how long it flies to its apogee (i.e.apex). But the a that applies here is the acceleration due to gravity. It is acceleration, but we just call it g: t = v/g = 7m/s / 9.8m/s^2 = .714 s.
Now let's put that time into equation 1 to get free flight distance: s = 1/2 gt^2 = 1/2 * 9.8 * (.714)^2 = 2.5 m. Now it started at .7 m, so it gets 2.5 + .7 = 3.2m off the ground.
b) 3.2m off the ground is your answer.
Please check my arithmetic, and you should be all set.
1) s = distance = 0.5 a t^2
2) v = a * t
3) average v = v at end minus v at start, divided by 2.
The 35m/s^2 and the 0.7 m leap give enough information to solve the problem. First, you want to know what velocity the bok is going when its feet leave the ground.
The givens are acceleration and distance. Equation 1 uses those terms. But it needs to solve for t. If you do a little algebra, you will find t = sqrt(2s/a). Putting in our values, you get sqrt(2*0.7/35) = 0.2 sec. Thinking about it, that sounds just about right.
Why do I use 35m.s^2 as upward acceleration, but ignore gravity? Because the problem states it is accelerating upward at that rate. Gravity could be anything or the bok could be out in space under no gravity, but it would still spring at that acceleration and its feet would leave the surface in 0.2 sec. In space, it would have to press less hard to accelerate at 35m/s^2 than it would off the earth.
Now the next phase is free flight. You want to know the velocity. Use equation 3 (works when a =constant): V = a*t = 35m/s^2 * 0.2s = 7 m/s.
a) 7 m/s is your answer for your a).
Ok, it is traveling up at 7 m/s, and starts its free flight 0.7 meters off the ground. Use equation 2 to find out how long it flies to its apogee (i.e.apex). But the a that applies here is the acceleration due to gravity. It is acceleration, but we just call it g: t = v/g = 7m/s / 9.8m/s^2 = .714 s.
Now let's put that time into equation 1 to get free flight distance: s = 1/2 gt^2 = 1/2 * 9.8 * (.714)^2 = 2.5 m. Now it started at .7 m, so it gets 2.5 + .7 = 3.2m off the ground.
b) 3.2m off the ground is your answer.
Please check my arithmetic, and you should be all set.