A stone is thrown vertically upward with a speed of 49 ms-!. Then the velocity of the stone, one second
before it reaches the maximum height is
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1
Answer:
No
Explanation:
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Answered by
4
Explanation:
Initial velocity (u) = 49 m/s
Let the time taken to reach the maximum height be t.
Acceleration due to gravity = - 9.8 m/s^2
Final velocity = 0 m/s
Now, using first equation of motion:
v_f = v_i + atv
f
=v
i
+at
0 = 49 - 9.8 × t
t = \frac{49}{9.8}t=
9.8
49
t = 5 seconds
Now, we need to find the velocity just 1 second before reaching the highest point.
Hence, time = 4 seconds
v_f = v_i + atv
f
=v
i
+at
v_f = 49 - 9.8 \times 4v
f
=49−9.8×4
v_f = 9.8 \frac{m}{s}v
f
=9.8
s
m
Hence, the velocity just 1 second before reaching the highest point is 9.8 m/s.
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