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accelerationproblemusingaslingshotakidshootsarockstraightupat30m/sfromthetopoftheRogunDaminTajikistan,theworld'shighestdam.therockfinallystrikesthewater325mbelowitsstartingpoint.Assumethefaceofthedamisvertical
题目详情
acceleration problem
using a slingshot a kid shoots a rock straight up at 30m/s from the top of the Rogun Dam in Tajikistan,the world's highest dam.the rock finally strikes the water 325m below its starting point.
Assume the face of the dam is vertical.
1.how high does the rock rise?
2.how long is the rock in the air?
3.how long would it have been in the air if it were launched staright down?
4.how long would it have been the air if it were dropped?
using a slingshot a kid shoots a rock straight up at 30m/s from the top of the Rogun Dam in Tajikistan,the world's highest dam.the rock finally strikes the water 325m below its starting point.
Assume the face of the dam is vertical.
1.how high does the rock rise?
2.how long is the rock in the air?
3.how long would it have been in the air if it were launched staright down?
4.how long would it have been the air if it were dropped?
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答案和解析
This can not be so hard a problem.
1、The rock first moves upward until it reachs the highest point ,then fall down.
It's initial antrorse speed of the rock is 30m/s while it's acceleration is just the acceleration of the gravity.
According to the equation V2^2 - V1^2 = 2as
while the V2 = 0 ,V1=30m/s and a = -g = -10m/s^2
So ,the hight of the rock moving up is s = 30*30/(2*10) = 45m
2、As we know ,the whole progress can be divided into 2 parts.
At first ,the rock run up until its speed is 0 .
The time is t1 ,so t1 = V2-V1/(-g) = 3s .
Then the rock act free-fall motion while the hight is 325 + 45 = 370m.According to the equation s = 0.5gt^2 ,we assume that the time is t2 ,so t2 = √（2*370/10） = 8.6s .
Finally the whole time is T = t1 + t2 = 11.6s
3、If it were launched staright down ,the rock has an downward initial speed of 30m/s while the acceleration of gravity acts.
According to the equation s = V1*t + 0.5gt^2
while s = 325m ,V1 = 30m/s and g = 10m/s^2
So the key to the equation is the solve of the proble.
t = √（74）- 3 = 5.6s
4、To be honest ,I can not sure the meaning of the word “dropped”.
Because there is no angle we known ,so I can only assume that the initial speed is horizontal.
So the time of the rock running in the air depend on the time of motion of the vertical free-fall motion.
The time is t and the equation is s = 0.5gt^2
while s = 325m ,g = 10m/s^2
so t = √65 = 8.06s
1、The rock first moves upward until it reachs the highest point ,then fall down.
It's initial antrorse speed of the rock is 30m/s while it's acceleration is just the acceleration of the gravity.
According to the equation V2^2 - V1^2 = 2as
while the V2 = 0 ,V1=30m/s and a = -g = -10m/s^2
So ,the hight of the rock moving up is s = 30*30/(2*10) = 45m
2、As we know ,the whole progress can be divided into 2 parts.
At first ,the rock run up until its speed is 0 .
The time is t1 ,so t1 = V2-V1/(-g) = 3s .
Then the rock act free-fall motion while the hight is 325 + 45 = 370m.According to the equation s = 0.5gt^2 ,we assume that the time is t2 ,so t2 = √（2*370/10） = 8.6s .
Finally the whole time is T = t1 + t2 = 11.6s
3、If it were launched staright down ,the rock has an downward initial speed of 30m/s while the acceleration of gravity acts.
According to the equation s = V1*t + 0.5gt^2
while s = 325m ,V1 = 30m/s and g = 10m/s^2
So the key to the equation is the solve of the proble.
t = √（74）- 3 = 5.6s
4、To be honest ,I can not sure the meaning of the word “dropped”.
Because there is no angle we known ,so I can only assume that the initial speed is horizontal.
So the time of the rock running in the air depend on the time of motion of the vertical free-fall motion.
The time is t and the equation is s = 0.5gt^2
while s = 325m ,g = 10m/s^2
so t = √65 = 8.06s
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