A boy drops a stone from the top of tower of height 4.9m. What is the velocity with which the ball hits the ground?
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the initial velocity of stone = 0m/s
distance = 4.9m
acceleration due to gravity = 9.8m/s
according to equation of motion
s= ut+1/2at²
4.9=0×t + 1/2 ×9.8 t²
4.9= 4.9t²
t²= 1
t=1
now,
v= u+at
v= 0+ 9.8×1
therefore velocity equals 9.8ms-1
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hope it helps....
distance = 4.9m
acceleration due to gravity = 9.8m/s
according to equation of motion
s= ut+1/2at²
4.9=0×t + 1/2 ×9.8 t²
4.9= 4.9t²
t²= 1
t=1
now,
v= u+at
v= 0+ 9.8×1
therefore velocity equals 9.8ms-1
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hope it helps....
seenu21:
you answer is correct but it can find by using v² - u² =2as
yes..i used that eq coz i. forgot the ques and found t..
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