write an expression in terms of velocity initial, theta and g for time it takes to travel max height
The correct answer and explanation is:
Expression for Time to Reach Maximum Height
The time tt it takes for a projectile to reach maximum height is given by the following expression: t=visin(θ)gt = \frac{v_i \sin(\theta)}{g}
Where:
- viv_i is the initial velocity of the projectile
- θ\theta is the angle of projection above the horizontal
- gg is the acceleration due to gravity (approximately 9.8 m/s29.8 \, \text{m/s}^2 on Earth)
Explanation (300 words)
In projectile motion, the time it takes to reach the maximum height is the moment when the vertical component of the velocity becomes zero. At launch, the object is projected with an initial velocity viv_i at an angle θ\theta above the horizontal. This velocity can be broken down into two components:
- Horizontal component: vicos(θ)v_i \cos(\theta) – constant throughout flight (ignoring air resistance)
- Vertical component: visin(θ)v_i \sin(\theta) – changes due to gravity
As the projectile rises, gravity acts downward, reducing the vertical component of velocity. Eventually, the vertical velocity becomes zero—this is the highest point (maximum height).
To find the time to reach this point, we use the following kinematic equation for vertical motion: vf=vi+atv_f = v_i + a t
At maximum height, the final vertical velocity vf=0v_f = 0. The acceleration aa is gravity, but acting downward, so a=−ga = -g. The initial vertical velocity is visin(θ)v_i \sin(\theta). Plugging in: 0=visin(θ)−gt0 = v_i \sin(\theta) – g t
Solving for tt: t=visin(θ)gt = \frac{v_i \sin(\theta)}{g}
This equation tells us that the time to reach max height depends directly on the initial vertical velocity and inversely on gravity. A greater angle θ\theta (closer to vertical) increases the vertical component of velocity, hence increasing time to max height. A larger gravitational force reduces that time. This formula is essential in physics and engineering for predicting projectile behavior.