Question

A ball will be thrown upward from a height of 20 meters above the ground with an initial velocity of 15 meters per second. Its distance from the ground, d, is a function of the time in seconds since the ball was thrown, t. This function is graphed below. At 3 seconds, how high will the ball be off the ground?

Answer 1

In physics, the vertical height of a projectile at a given time can be calculated with a kinematic equation, with variables including initial height, initial velocity, time, and acceleration due to gravity.

Step-by-step explanation:

The question asks about the height of a ball thrown upward from a given height with an initial velocity after a certain amount of time. This scenario involves projectile motion, a common topic in physics. The height of the ball at 3 seconds can be calculated using the kinematic equations for projectile motion, specifically considering vertical motion. The formula h(t) = h0 + v0t + ½at², where h0 is the initial height, v0 is the initial velocity, a is the acceleration due to gravity (which is -9.8 m/s²), and t is the time in seconds. However, the question assumes a given graph from which the information is to be extracted, but the graph is not provided.