A grenade is thrown up into the air with an initial velocity of 27 m/s. How long does it take to return to the thrower's hand?
I got 5.4 s, but I am very unsure.
well, you know that
y(t) = 27t - 4.9t^2
y=0 when
27t-4.9t^2 = 0
t = 27/4.9 = 5.51
You are close, but I wonder how you got what you did...
Ah. I see you used g = 10 m/s^2, not 9.8
OK, you are correct.
To determine how long it takes for the grenade to return to the thrower's hand, we can use the concept of projectile motion.
First, we need to recognize that throwing the grenade up into the air constitutes vertical motion, disregarding any horizontal motion. Since we are only interested in the time it takes for the grenade to return to the thrower, we can ignore factors such as air resistance.
The motion of the grenade can be modeled using the equations of motion:
1. The vertical displacement of an object in projectile motion can be found using the formula:
Δy = V₀y * t + (1/2) * g * t²
where:
Δy is the vertical displacement (which is zero for a projectile that returns to its initial height)
V₀y is the initial vertical velocity (in this case, 27 m/s)
t is the time taken
g is the acceleration due to gravity (approximately -9.8 m/s²)
2. The equation of motion for vertical velocity is:
Vf = V₀y + g * t
where:
Vf is the final vertical velocity (which is zero when the grenade returns to the thrower's hand)
Using these two equations, we can find the time it takes for the grenade to return to the thrower:
Step 1: Setting the initial vertical velocity equal to zero:
0 = 27 m/s - 9.8 m/s² * t
Step 2: Solving for time:
9.8 m/s² * t = 27 m/s
t = 27 m/s / 9.8 m/s²
t ≈ 2.76 seconds
Therefore, it takes approximately 2.76 seconds for the grenade to return to the thrower's hand.