- #1
faisal
- 46
- 0
''A DART LEAVES THE THROWER HAND HORIZONTALLY AT A height of 1.9m above the ground, calculate the time taken by the dart to reach the board and the horizontal velocity at which the dart left the throwers hand.''
this is what iv worked out
x-0.4
u-?
v-0
a-0
t-?
i than went onto finding the vertical acceleration
x-0.4, u-?, v-0. a-9.81, t-?
v^2=u^2+2ax
v^2-u^2=2ax
v^2-v^2-u^2=2ax-v^2
u^2=2ax-v^2
-u^2=2ax-v^2
2x9.81x0.4=7.84
i than found the square root of 7.84 however it was wrong, since it was -u^2
this is what iv worked out
x-0.4
u-?
v-0
a-0
t-?
i than went onto finding the vertical acceleration
x-0.4, u-?, v-0. a-9.81, t-?
v^2=u^2+2ax
v^2-u^2=2ax
v^2-v^2-u^2=2ax-v^2
u^2=2ax-v^2
-u^2=2ax-v^2
2x9.81x0.4=7.84
i than found the square root of 7.84 however it was wrong, since it was -u^2