Question

A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a height of 2.00 m above the ground with a speed of 16.0 m/s. How long does the ball remain in the air before striking the ground?

Answer 1

To determine the time a projectile remains in the air, calculate the initial vertical velocity using the sine of the launch angle and then apply kinematic equations to solve for time using the vertical displacement and acceleration due to gravity.

Step-by-step explanation:

The student's question is related to the time a projectile remains in the air, which is a topic in Physics, specifically in kinematics. To find the time of flight for the projectile, we need to consider the vertical motion since the horizontal and vertical motions are independent.

Given that the initial speed of the ball is 16.0 m/s and it is thrown at a 60.0° angle, we can find the initial vertical velocity (Vy) using the sine function: Vy = 16.0 m/s * sin(60.0°).

The vertical displacement Δy for the ball when it hits the ground will be -2.00 m (negative since it's ending below the point of release). We will use the following kinematic equation for vertical displacement: Δy = Vy * t + 0.5 * g * t^2, where g is the acceleration due to gravity (-9.8 m/s^2, negative as it's directed downwards).

By substituting the known values and solving the quadratic equation for t (time), we can determine the total time the ball remains in the air.

Answer 2

Im going to give you the formula, so that you can replace the data with the correspondent variables. The formula is :

y(t)=y0+v0sin(θ)t−(1/2)gt^2

Remember that all objects accelerate uniformly near the surface of the Earth. Therefore, the time that the ball will hit the ground is determined by the equation i gave you. Hope this can help you