Numerical_A ball is thrown vertically upward with velocity of 10m/s. calculate the height attained by the ball and the time taken by it to reach the highest point if the acceleration of the ball is -10m/s^2 during its motion.
Answers
Answer: The motion equations applicable for an object thrown upward are:
During upward movement:
V = U – gt………(i)
H =Ut – (1/2) g t^2…….(ii)
V^2 = U^2 – 2gH…..(iii)
During downward movement:
V = U + gt………(iv)
H =Ut + (1/2) g t^2…….(v)
V^2 = U^2 + 2gH…..(vi)
List of formulas related to a ball thrown vertically upward [formula set]
If a ball is thrown vertically upwards with an initial velocity V0 then here is a set of formula for your quick reference.
1) Maximum height reached =
H = V02 / (2 g)
2) Velocity at the highest point = 0
3) Time for upward movement = V0 /g
4) Time for downward movement =
V0 /g
5) Total time of travel in air = (2 V0 )/g
6) Acceleration of the ball = acceleration due to gravity (g) acting downwards, towards the center of earth [ignoring air resistance]
7) Forces acting on the ball = Gravity (gravitational force exerted by the earth)
[ignoring air resistance]
Problem-solving using these formulas
A ball is thrown vertically upwards with a velocity of 49m/s calculate maximum height and time taken to reach maximum height.
H = U2/(2g) = (492)/(2 x 9.8)=122.5 m
T = U/g = 49/9.8 = 5 sec
A ball is thrown vertically upward with a velocity of 20 m/s. calculate maximum height and time taken to reach maximum height.
H = U2/(2g) = (202)/(2 x 9.8)=20.4 m
T = U/g = 20/9.8 = 2.04 sec