A stone is thrown with a velocity of 15 m/s at an angle of 60° to the horizontal . calculate
(a) the time it takes before the stone hits the ground
v = 15 sin60° - 9.8t
find t when v=0
that is when it is at the top of its arc, so double that time for it to fall back down.
To calculate the time it takes before the stone hits the ground, we can break the initial velocity of the stone into its horizontal and vertical components.
Given:
Initial velocity (v) = 15 m/s
Launch angle (θ) = 60°
The vertical component of the initial velocity can be calculated using the formula:
v_vertical = v * sin(θ)
v_vertical = 15 * sin(60°)
v_vertical ≈ 12.99 m/s
The time it takes for the stone to reach the highest point in its trajectory can be calculated using the formula:
Time to reach highest point (t1) = v_vertical / g
where g is the acceleration due to gravity (approximately 9.8 m/s²)
t1 = 12.99 / 9.8
t1 ≈ 1.32 seconds
Since the stone takes the same amount of time to reach its maximum height as it does to fall back down to the ground, the total time it takes for the stone to hit the ground can be found by doubling t1.
Total time (t_total) = 2 * t1
t_total ≈ 2.64 seconds
Therefore, it takes approximately 2.64 seconds for the stone to hit the ground.