Answer:
a) - 9.8 m/s²
b) 0 m/s
c) 9.8 m/s
d) 4.9 m
Explanation:
(a)
While the ball is in flight, acceleration due to gravity acts on it. hence
a = acceleration due to gravity = - 9.8 m/s²
The negative sign indicates downward direction
b)
At the maximum height, the ball comes to a momentary stop. hence the velocity of the ball at the maximum height is zero.
0 m/s
c)
Consider the motion of the ball
v₀ = initial velocity of the ball at the time of throw = v
a = acceleration due to gravity = - 9.8 m/s²
t = time interval = 2 s
v = final velocity just before he catch the ball = - v
using the kinematics equation
v = v₀ + a t
- v = v + (- 9.8) (2)
v = 9.8 m/s
d)
h = maximum height reached
v₀ = initial velocity of the ball at the time of throw = 9.8 m/s
v = final velocity of the ball at the highest point = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
using the equation
v² = v₀² + 2 a h
0² = 9.8² + 2 (- 9.8) h
h = 4.9 m
