A grenade thrown straight up in the air has an initial speed of 27 m/s. How long will it take before it returns to the thrower's hand?
To find the time it takes for the grenade to return to the thrower's hand, we can use the equation of motion for objects thrown vertically:
h = ut - (1/2)gt^2
Where:
h is the maximum height the grenade reaches
u is the initial speed of the grenade
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken
When the grenade reaches its maximum height, its velocity becomes zero before it starts to descend. At this point, its displacement (h) is equal to its initial height (0) as it is thrown straight up. Therefore:
0 = (27)t - (1/2)(9.8)t^2
Rearranging the equation:
4.9t^2 - 27t = 0
Now, we can solve this quadratic equation to find the time it takes for the grenade to return to the thrower's hand.
Dividing the equation by t:
4.9t - 27 = 0
Simplifying:
4.9t = 27
t = 27 / 4.9
t ≈ 5.51 seconds
Therefore, it will take approximately 5.51 seconds for the grenade to return to the thrower's hand.