Question

A projectile is thrown with an initial speed vo = 25.0 m/s at an initial angle with the horizontal ao = 45.0. a . Find the time T when the projectile is at its maximum height. b. Find the position (x, y) at the maximum height c. At time T, find the components of the acceleration vector and the velocity vector. d. Sketch the trajectory of the projectile. On your sketch, label the position of the projectile at time T, draw the velocity vector, and draw the components of the acceleration vector.

Answer 1

Answer:

See the explanation below

Explanation:

To solve this problem we must decompose the initial speeds into x & y.

$v_{o}_{x}=25*cos(45)=17.67[m/s]\\v_{o}_{y}=25*sin(45)=17.67[m/s]$

The acceleration of  gravity is equal to g = 9.81[m/s^2] downward.

The maximum height is when the velocity of the projectile is zero in the component y, that is, it will not be able to go higher, by means of the following kinematic equation we can find that time, for that specific condition.

a)

$v_{y}=(v_{y})_{0}+a*t\\0 = 17.67 - 9.81*t\\17.67 = 9.81*t\\t=1.8 [s]$

Note: Acceleration is taken as negative as it is directed downwards.

b)

The position in the x component can be found using the following kinematic equation

$x=(v_{x})_{o}*t\\x=17.67*1.8\\x=31.82[m]$

The position in the y component can be found using the following kinematic equation

$y =(v_{y})_{o}*t+\frac{1}{2} *g*t^{2} \\y=17.67*1.8-0.5*9.81*(1.8)^{2}\\ y=15.91[m]$

c)

Since the motion on the X-axis is at constant speed, there is no acceleration, so the only acceleration that exists is due to gravity

d)

In the attached image we can see, the projectile with the vectors of acceleration and velocity.