A ball is projected vertically upward from the ground. The ball is at same height at t=3s and t=9s the height of ball at t=5s is......plz reply fast.....
t = 6 s is the apex of the flight
max height = 1/2 g 6^2
the ball is at the same height at t = 5 and t = 7
... 1 second from max ... 4.9 m below max
To find the height of the ball at t=5s, we need to use the kinematic equation for vertical motion:
h = v₀t + (1/2)gt²
where:
h is the height of the ball
v₀ is the initial velocity of the ball
g is the acceleration due to gravity (-9.8 m/s²)
t is the time
Let's break down the problem step by step:
Step 1: Find the initial velocity
Since the ball is projected vertically upward from the ground, its initial velocity is the velocity at t=0s. Unfortunately, the problem doesn't provide the initial velocity, so we can't calculate the exact height of the ball. We need more information.
Step 2: Identify the heights at t=3s and t=9s
The problem states that the ball is at the same height at t=3s and t=9s. This implies that the ball reaches its maximum height and starts to descend between these two times. However, given that information alone, we still cannot determine the height at t=5s.
In order to find the height at t=5s, we need the initial velocity of the ball or more information about its motion. Without that information, it is not possible to calculate the height at t=5s based on the given data.