Question

A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air​ resistance, its height in feet t seconds after launch is given by s equals negative 16 t squared plus v0 ts=−16t2+v0t. Find the​ time(s) that the projectile will​ (a) reach a height of 240 ft and​ (b) return to the ground when v0 equals=128 feet per second.
​(a) Find the​ time(s) that the projectile will reach a height of 240 ft when v0 equals=128 feet per second. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

Answer 1

This problem involves the use of a kinematic equation since it involves the motion of an object. The equation for the height of the object is given as:

s = -16*t^2+v_o*t

Also, the initial velocity, v_o, was also said to be equal to 128 ft/s.

For the first question, the​ time(s) that the projectile will reach a height of 240 ft when v_o is 128 feet per second, can be solved using the given equation and the quadratic formula. The resulting quadratic equation is then -16*t^2 + 128*t -240 =0, where a =16, b =128, and c =-240. The quadratic formula is equal to [-b ± sqrt(b^2 -4ac)]/2a. This gives two answers t = 3 seconds and t = 5 seconds. This might be because the projectile has a parabolic path, thus, it reaches the height of 240 ft, before and after it reaches a peak.

For the second question, the time it takes for the projectile to reach the ground is obtained by setting the distance, s, equal to zero. In this case, t = 8 seconds.