Answer:
a) 0.76 m
b) 1.53 s
c) 0.39 s
d) 3.8 m/s
Explanation:
This is a problem in which we need to use the equations pertaining Uniformly Accelerated Motion, as the acceleration during all this process is constant: The gravitational pull on the ball, 9.8 m/s².
To make things easier, we can divide this process in two parts: The first one (A) is from the moment the ball is thrown, until the moment it reaches it highest point and momentarily stops. The second one (B) from the moment it starts descending until it hits the ground.
a) During part A, we use the formula Vf²=Vi² +2*a*x . Where Vf is the final velocity (0 m/s, as the ball stopped in midair), Vi is the initial velocity (15 m/s), a is the acceleration (-9.8 m/s², it has a minus sign, as it goes against the direction of the movement) and x is the distance; thus, we're left with:
0 m/s=(15.0 m/s)²+2*(-9.8 m/s²)*x
We solve for x
x = 0.76 m
b) The formula is Vf=Vi +a*t, where t is the time. We're left with:
0 m/s=15.0 m/s + (-9.8 m/s²)*t
We solve for t
t= 1.53 s
c) Now we focus on part B, and use the formula x=Vi * t + [tex]\frac{at^{2}}{2}[/tex] , with the difference that the Vi is 0 m/s. We already know the value of x in exercise a). Note that a does not have a negative sign, as the direction of movement is opposite to the direction of part A
0.76 m=0 m/s * t +[tex]\frac{9.8\frac{m}{s^{2} } *t^{2} }{2}[/tex]
Solve for t
0.76=4.9t²
t=0.39 s
d) Once again we use the formula Vf=Vi +a*t, using the value of t previously calculated in exercise c).
Vf=0 m/s + 9.8 m/s² * 0,39 s
Vf=3.8 m/s
