A particle is projected vertically upwards from the ground with speed [tex]$30 ms^{-1}$[/tex]. Calculate the:
(a) Maximum height reached by the particle.
(b) Time taken by the particle to return to the ground.
(c) Time (s) taken for the particle to attain a height of 4 cm above the ground. Take [tex]$g = 10 ms^{-2}$[/tex]
Answer (1)
The maximum height is found using v 2 = u 2 + 2 a s , resulting in s = 2 g u 2 = 2 × 10 3 0 2 = 45 m.
The time to return to the ground is calculated using s = u t + 2 1 a t 2 , giving t = g 2 u = 10 2 × 30 = 6 s.
The time to reach 4 m is found by solving the quadratic equation 5 t 2 − 30 t + 4 = 0 , resulting in two times: t = 3 ± 5 205 , approximately 0.136 s and 5.864 s.
The maximum height reached by the particle is 45 m , the time taken by particle to return to the ground is 6 s and the time taken for the particle to attain a height of 4 m above the ground is 0.136 s and 5.864 s .
Explanation
Problem Setup We are given that a particle is projected vertically upwards from the ground with an initial speed of 30 m s − 1 . We need to find (a) the maximum height reached by the particle, (b) the time taken for the particle to return to the ground, and (c) the time taken for the particle to attain a height of 4 m above the ground. We are also given that the acceleration due to gravity is g = 10 m s − 2 .
Calculating Maximum Height At the maximum height, the final velocity of the particle will be v = 0 m s − 1 . We can use the following kinematic equation to find the maximum height s :
v 2 = u 2 + 2 a s
where u = 30 m s − 1 is the initial velocity, a = − g = − 10 m s − 2 is the acceleration due to gravity (negative since it opposes the motion), and v = 0 m s − 1 is the final velocity at the maximum height. Plugging in the values, we get:
0 2 = ( 30 ) 2 + 2 ( − 10 ) s
0 = 900 − 20 s
20 s = 900
s = 20 900 = 45 m
Therefore, the maximum height reached by the particle is 45 meters.
Calculating Time to Return to Ground To find the time taken for the particle to return to the ground, we can use the following kinematic equation:
s = u t + 2 1 a t 2
where s = 0 m (since the particle returns to the ground), u = 30 m s − 1 is the initial velocity, and a = − g = − 10 m s − 2 is the acceleration due to gravity. Plugging in the values, we get:
0 = 30 t + 2 1 ( − 10 ) t 2
0 = 30 t − 5 t 2
0 = t ( 30 − 5 t )
This gives us two possible solutions for t : t = 0 s (which is the initial time when the particle is projected) and 30 − 5 t = 0 , which gives t = 5 30 = 6 s . Therefore, the time taken for the particle to return to the ground is 6 seconds.
Calculating Time to Reach 4m Height To find the time taken for the particle to attain a height of 4 m above the ground, we can use the same kinematic equation:
s = u t + 2 1 a t 2
where s = 4 m , u = 30 m s − 1 is the initial velocity, and a = − g = − 10 m s − 2 is the acceleration due to gravity. Plugging in the values, we get:
4 = 30 t + 2 1 ( − 10 ) t 2
4 = 30 t − 5 t 2
Rearranging the equation, we get a quadratic equation:
5 t 2 − 30 t + 4 = 0
We can solve for t using the quadratic formula:
t = 2 a − b ± b 2 − 4 a c
where a = 5 , b = − 30 , and c = 4 . Plugging in the values, we get:
t = 2 ( 5 ) 30 ± ( − 30 ) 2 − 4 ( 5 ) ( 4 )
t = 10 30 ± 900 − 80
t = 10 30 ± 820
t = 10 30 ± 2 205
t = 3 ± 5 205
So, we have two possible values for t :
t 1 = 3 + 5 205 ≈ 3 + 5 14.3178 ≈ 3 + 2.86356 ≈ 5.86356 s
t 2 = 3 − 5 205 ≈ 3 − 5 14.3178 ≈ 3 − 2.86356 ≈ 0.13644 s
Therefore, the particle attains a height of 4 m at approximately t = 0.13644 s (on the way up) and t = 5.86356 s (on the way down).
Final Answer The maximum height reached by the particle is 45 meters. The time taken for the particle to return to the ground is 6 seconds. The time taken for the particle to attain a height of 4 m above the ground is approximately 0.13644 seconds (on the way up) and 5.86356 seconds (on the way down).
Examples
Understanding projectile motion is crucial in various real-world applications. For example, engineers use these principles to design ballistics for artillery, ensuring that projectiles reach their intended targets accurately. Similarly, in sports, athletes and coaches use projectile motion concepts to optimize performance, such as determining the best angle and initial velocity for throwing a ball or jumping.
