A ball is dropped from rest from the top of a tower. How far will it fall in 4 seconds? (Take g = 10 m/s²)
Explanation
Since the ball is dropped from rest, u = 0
Using s = ut + ½gt²
s = (0 × 4) + (½ × 10 × 4²)
s = 0 + 80
s = 80 m
✓
Conclusion
The ball falls 80 m in 4 seconds.
2
A ball is thrown vertically downward from the top of a tower with an initial velocity of 10 m/s. What are its position (distance fallen from the starting point) and velocity after 8 seconds? (Take g = 10 m/s² and neglect air resistance.)
Explanation
Taking downward as the positive direction:
Position: s = ut + ½gt²
s = (10 × 8) + (½ × 10 × 8²)
s = 80 + 320 = 400 m downward
Velocity: v = u + gt
v = 10 + (10 × 8) = 90 m/s downward
✓
Conclusion
After 8 seconds, the ball is 400 m below its starting point and is moving at 90 m/s downward.
3
A person throws a ball vertically upward with an initial velocity of 70 m/s. Calculate the maximum height reached above the point of projection. Ignore air resistance. (Take g = 10 m/s²)
Explanation
At maximum height, final velocity v = 0
Using the equation of motion: v² = u² + 2as
Since a = −g, we get 0 = u² − 2gh
Therefore, h = u² / 2g
h = (70 × 70) / (2 × 10)
h = 245 m
✓
Conclusion
The maximum height reached by the ball is 245 m.
4
A person throws a ball vertically upward with an initial velocity of 100 m/s. How long does the ball remain in the air before returning to the person’s hand? Ignore air resistance. (Take g = 10 m/s²)
Explanation
At the highest point, final velocity v = 0
Using v = u − gt
Time to reach maximum height: t_up = u / g
For return to the same height, total time T = 2t_up
Therefore, T = 2u / g
T = (2 × 100) / 10
T = 20 s
✓
Conclusion
The ball remains in the air for 20 seconds.
5
A person throws a ball vertically upward with an initial velocity of 80 m/s. How much time does the ball take to reach maximum height, and what is its velocity when it returns to the thrower’s hand? Ignore air resistance. (Take g = 10 m/s².)
Explanation
Taking upward as the positive direction:
At maximum height, v = 0
Using v = u − gt
0 = u − gt
Time to maximum height: t = u / g
t = 80 / 10 = 8 s
When the ball returns to the same height, its speed equals its initial speed
Return velocity = −80 m/s (downward)
✓
Conclusion
Time to maximum height = 8 s; return velocity = −80 m/s (downward).
6
Two balls are thrown from the top of a cliff of height 34 m. One is thrown vertically upward and the other vertically downward, both with the same initial speed of 18 m/s. Both balls hit the ground below the cliff at different times. Which ball hits the ground at the greater speed? Ignore air resistance. (Take g = 10 m/s²)
Explanation
For the ball thrown upward, its initial velocity is +u if upward is positive.
For the ball thrown downward, its initial velocity is −u.
Using v² = u² + 2as, the final speed depends on the initial speed magnitude and vertical displacement.
For either ball: v² = (18)² + 2 × 10 × 34
v² = 1004
v = √1004 ≈ 31.69 m/s
Both balls have the same final speed; their flight times differ.
✓
Conclusion
Both balls hit the ground at the same speed, approximately 31.69 m/s.
7
A person standing at the edge of a cliff throws a ball vertically upward with an initial speed of 90 m/s. The base of the cliff is 50 m below the point of projection. Ignoring air resistance and taking g = 10 m/s², determine (a) the time taken by the ball to reach the base of the cliff and (b) the total distance travelled by the ball.
Explanation
Taking the launch point as y = 0 and upward as positive:
At the base of the cliff, displacement y = −h
Using y = ut − ½gt²
−50 = 90t − ½ × 10 × t²
Solving the quadratic equation gives the positive time:
t = [u + √(u² + 2gh)] / g = 18.54 s
Maximum height above the cliff edge: H = u² / 2g
H = (90)² / (2 × 10) = 405 m
Total distance = upward distance + downward distance
Total distance = H + (H + 50)
Total distance = 405 + 405 + 50 = 860 m
✓
Conclusion
(a) Time to reach the base = 18.54 s. (b) Total distance travelled = 860 m.