Question:
A student throws a ball straight up with an initial velocity of 25 m/s. The ball reaches its maximum height after 4 seconds and then falls back down. The acceleration due to gravity is 9.8 m/s².
(a) What is the maximum height reached by the ball? Use the equation of motion for vertical motion to solve this problem.
(b) How long does it take for the ball to reach the ground again? Use the equation of motion for vertical motion to solve this problem.
(c) What is the velocity of the ball just before it hits the ground? Use the equation of motion for vertical motion to solve this problem.
Solution:
(a) To find the maximum height reached by the ball, we can use the equation of motion for vertical motion:
v f 2 = v i 2 + 2 a Δ y v_f^2 = v_i^2 + 2a \Delta y v f 2 = v i 2 + 2 a Δ y Where:v f v_f v f v i v_i v i a a a Δ y \Delta y Δ y
Rearranging the equation and substituting the known values, we have:
0 2 = ( 25 m/s ) 2 + 2 ( − 9.8 m/s 2 ) Δ y 0^2 = (25 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2) \Delta y 0 2 = ( 25 m/s ) 2 + 2 ( − 9.8 m/s 2 ) Δ y Simplifying the equation:
0 = 625 m²/s² − 19.6 m/s² Δ y 0 = 625 \, \text{m²/s²} - 19.6 \, \text{m/s²} \Delta y 0 = 625 m²/s² − 19.6 m/s² Δ y 19.6 m/s² Δ y = 625 m²/s² 19.6 \, \text{m/s²} \Delta y = 625 \, \text{m²/s²} 19.6 m/s² Δ y = 625 m²/s² Δ y = 625 m²/s² 19.6 m/s² \Delta y = \frac{625 \, \text{m²/s²}}{19.6 \, \text{m/s²}} Δ y = 19.6 m/s² 625 m²/s² Δ y ≈ 31.89 m \Delta y \approx 31.89 \, \text{m} Δ y ≈ 31.89 m Therefore, the maximum height reached by the ball is approximately 31.89 m.
(b) To find the time it takes for the ball to reach the ground again, we can use the equation of motion for vertical motion:
Δ y = v i t + 1 2 a t 2 \Delta y = v_i t + \frac{1}{2} a t^2 Δ y = v i t + 2 1 a t 2 Where:Δ y \Delta y Δ y v i v_i v i a a a t t t
Rearranging the equation and substituting the known values, we have:
− 31.89 m = ( 25 m/s ) t + 1 2 ( − 9.8 m/s² ) t 2 -31.89 \, \text{m} = (25 \, \text{m/s}) t + \frac{1}{2} (-9.8 \, \text{m/s²}) t^2 − 31.89 m = ( 25 m/s ) t + 2 1 ( − 9.8 m/s² ) t 2 Simplifying the equation:
− 31.89 m = 25 m/s ⋅ t − 4.9 m/s² ⋅ t 2 -31.89 \, \text{m} = 25 \, \text{m/s} \cdot t - 4.9 \, \text{m/s²} \cdot t^2 − 31.89 m = 25 m/s ⋅ t − 4.9 m/s² ⋅ t 2 Rearranging and setting the equation equal to zero:
4.9 m/s² ⋅ t 2 − 25 m/s ⋅ t − 31.89 m = 0 4.9 \, \text{m/s²} \cdot t^2 - 25 \, \text{m/s} \cdot t - 31.89 \, \text{m} = 0 4.9 m/s² ⋅ t 2 − 25 m/s ⋅ t − 31.89 m = 0 Using the quadratic formula:
t = − b ± b 2 − 4 a c 2 a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} t = 2 a − b ± b 2 − 4 a c Where:a = 4.9 m/s² a = 4.9 \, \text{m/s²} a = 4.9 m/s² b = − 25 m/s b = -25 \, \text{m/s} b = − 25 m/s c = − 31.89 m c = -31.89 \, \text{m} c = − 31.89 m
Calculating the discriminant:
b 2 − 4 a c = ( − 25 m/s ) 2 − 4 ( 4.9 m/s² ) ( − 31.89 m ) = 2500 m²/s² + 6227.24 m²/s² = 8727.24 m²/s² b^2 - 4ac = (-25 \, \text{m/s})^2 - 4(4.9 \, \text{m/s²})(-31.89 \, \text{m}) = 2500 \, \text{m²/s²} + 6227.24 \, \text{m²/s²} = 8727.24 \, \text{m²/s²} b 2 − 4 a c = ( − 25 m/s ) 2 − 4 ( 4.9 m/s² ) ( − 31.89 m ) = 2500 m²/s² + 6227.24 m²/s² = 8727.24 m²/s² Simplifying:
t = − ( − 25 ) ± 8727.24 m²/s² 2 ( 4.9 m/s² ) t = \frac{-(-25) \pm \sqrt{8727.24 \, \text{m²/s²}}}{2(4.9 \, \text{m/s²})} t = 2 ( 4.9 m/s² ) − ( − 25 ) ± 8727.24 m²/s² t = 25 m/s ± 93.41 m/s 9.8 m/s² t = \frac{25 \, \text{m/s} \pm 93.41 \, \text{m/s}}{9.8 \, \text{m/s²}} t = 9.8 m/s² 25 m/s ± 93.41 m/s We take the positive value as time cannot be negative:
t = 25 m/s + 93.41 m/s 9.8 m/s² t = \frac{25 \, \text{m/s} + 93.41 \, \text{m/s}}{9.8 \, \text{m/s²}} t = 9.8 m/s² 25 m/s + 93.41 m/s t = 118.41 m/s 9.8 m/s² t = \frac{118.41 \, \text{m/s}}{9.8 \, \text{m/s²}} t = 9.8 m/s² 118.41 m/s t ≈ 12.07 s t \approx 12.07 \, \text{s} t ≈ 12.07 s Therefore, it takes approximately 12.07 seconds for the ball to reach the ground again.
(c) To find the velocity of the ball just before it hits the ground, we can use the equation of motion for vertical motion:
v f = v i + a t v_f = v_i + a t v f = v i + a t Where:v f v_f v f v i v_i v i a a a t t t
Substituting the known values into the equation, we have:
v f = 25 m/s + ( − 9.8 m/s² ) ⋅ ( 12.07 s ) v_f = 25 \, \text{m/s} + (-9.8 \, \text{m/s²}) \cdot (12.07 \, \text{s}) v f = 25 m/s + ( − 9.8 m/s² ) ⋅ ( 12.07 s ) v f = 25 m/s − 118.216 m/s v_f = 25 \, \text{m/s} - 118.216 \, \text{m/s} v f = 25 m/s − 118.216 m/s v f ≈ − 93.22 m/s v_f \approx -93.22 \, \text{m/s} v f ≈ − 93.22 m/s Therefore, the velocity of the ball just before it hits the ground is approximately -93.22 m/s. (The negative sign indicates downward direction.)