1) An object is shot upward at 34.3 m/s. How high does it go? a) 60.0 m b) 180 m

Question

1) An object is shot upward at 34.3 m/s. How high does it go? a) 60.0 m
b) 180 m
c) 103 m
d) 240 m

2) How much time does it take the object in Problem 1 to reach the highest position? 1) An object is shot upward at 34.3 m/s. How high does it go? a) 60.0 m
b) 180 m
c) 103 m
d) 240 m

1) An object is shot upward at 34.3 m/s. How high does it go? 2) How much time does it take the object in Problem 1 to reach the highest position? a) 7.50 s
b)3.50 s
c)1.75 s d)1.87 s
3) What is the acceleration of the object in Problem 1 when it is shot upward initially? a) An object is shot upward at 34.3 m/s. How high does it go? 60.0 m 180 m 103 m 240 m How much time does it take the object in Problem 1 to reach the highest position? 7.50 s 3.50 s 1.75 s 1.87 s What is the acceleration of the object in Problem 1 when it is shot upward initially? a) 34.3 m/s2 uparrow 0 9.8 m/s3 uparrow 9.8 m/s2 downarrow 34.3 m/s2 downarrow

Explanation / Answer

answer #2 first and find the time by equation v(final)=v(initial)+a*t--->t=v(initial)/a because there is no velocity at the instant the object peaks. Therefore t=34.3/9.81=3.49 s. So your answer is b)

#1 is a) using y(final)-y(initial)=0.5*[v(initial)+v(final)]*t. You have time and y(initial)=0 so y(final)=0.5*[34.3]*3.49=60.028 m

#3 there is no work. You have to know that a projectile is only experiencing the acceleration of gravity which is 9.81 m/s^2 downward (in this case). Therefore your answer is d)

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